Kinetic energy, potential energy and mechanical energy

One formula or one step of reasoning. Core exercises first, the rest is extra practice.

1. Exercise 10-1Core

   The mass of a passenger car is $1200\,\text{kg}$, and it drives at a speed of $20.0\,\text{m/s}$.

   How great is the kinetic energy of the car? Answer in kilojoules.

   kJCheck
   Answer

   $240\,\text{kJ}$

   Show the solution

   Given

   mass $m = 1200\,\text{kg}$, speed $v = 20.0\,\text{m/s}$

   Find

   the kinetic energy $E_\text{k}$ of the car in kilojoules

   Principle

   A moving object has kinetic energy, which is proportional to the mass and the square of the speed. The speed and the mass of the car are known, so the kinetic energy is obtained directly.

   Formula

   $E_\text{k} = \tfrac{1}{2}\,mv^2$. The formula is already solved for the quantity asked.

   Conversions

   The mass and the speed are in SI units, so the result comes in joules. Finally $1\,\text{kJ} = 1000\,\text{J}$.

   Substitution

   $E_\text{k}$$= \tfrac{1}{2} \cdot 1200\,\text{kg} \cdot (20.0\,\text{m/s})^2$

   $= 600\,\text{kg} \cdot 400\,\text{m}^2/\text{s}^2$

   $= 240\,000\,\text{J}$

   $= 240\,\text{kJ}$

   Answer

   $240\,\text{kJ}$

   Two significant figures like the mass $1200\,\text{kg}$.

   Reasonableness

   With rounded numbers

   $\tfrac{1}{2} \cdot 1000\,\text{kg} \cdot 400\,\text{m}^2/\text{s}^2$

   $= 200\,\text{kJ}$

   so the order of magnitude is right. The speed $20\,\text{m/s} = 72\,\text{km/h}$ is a normal speed on a country road, and the kinetic energy of a car is then hundreds of kilojoules.

2. Exercise 10-2Core

   A flower pot with a mass of $2.50\,\text{kg}$ is on a balcony $8.00\,\text{m}$ above the ground.

   How great is the potential energy of the pot when the zero level is the ground?

   JCheck
   Answer

   $196\,\text{J}$

   Show the solution

   Given

   mass $m = 2.50\,\text{kg}$, height $h = 8.00\,\text{m}$ above the zero level, that is, the ground, acceleration of free fall $g = 9.81\,\text{m/s}^2$

   Find

   the potential energy $E_\text{p}$ of the pot in joules

   Principle

   Gravitational potential energy depends on the mass and the height above the chosen zero level. The difference in height is small, so $g$ is constant and $E_\text{p} = m g h$ holds.

   Formula

   $E_\text{p} = m g h$. The formula is already solved for the quantity asked.

   Conversions

   All quantities are in SI units, and $\text{kg} \cdot \text{m/s}^2 \cdot \text{m} = \text{J}$. No conversions are needed.

   Substitution

   $E_\text{p}$$= 2.50\,\text{kg} \cdot 9.81\,\text{m/s}^2 \cdot 8.00\,\text{m}$

   $= 196.2\,\text{J}$

   Answer

   $196\,\text{J}$

   Three significant figures like the mass and the height.

   Reasonableness

   With rounded numbers $2.5\,\text{kg} \cdot 10\,\text{m/s}^2 \cdot 8\,\text{m} = 200\,\text{J}$. The weight of the pot is about $25\,\text{N}$, and lifting it to eight metres takes an energy of a few hundred joules. The result is positive, because the pot is above the zero level.

3. Exercise 10-3

   A gull with a mass of $1.20\,\text{kg}$ flies at a speed of $10.0\,\text{m/s}$ at a height of $20.0\,\text{m}$ above sea level.

   How great is the mechanical energy of the gull when the zero level is sea level?

   JCheck
   Answer

   $295\,\text{J}$

   Show the solution

   Given

   mass $m = 1.20\,\text{kg}$, speed $v = 10.0\,\text{m/s}$, height $h = 20.0\,\text{m}$ above the zero level, that is, sea level, $g = 9.81\,\text{m/s}^2$

   Find

   the mechanical energy $E_\text{mech}$ of the gull in joules

   Principle

   Mechanical energy is the sum of kinetic energy and potential energy. Kinetic energy is obtained from the speed and potential energy from the height above sea level.

   Formula

   $E_\text{mech} = E_\text{k} + E_\text{p}$, so $E_\text{mech} = \tfrac{1}{2}\,mv^2 + m g h$

   Conversions

   All quantities are in SI units, so both terms come in joules. No conversions are needed.

   Substitution

   $E_\text{mech}$$= \tfrac{1}{2} \cdot 1.20\,\text{kg} \cdot (10.0\,\text{m/s})^2 + 1.20\,\text{kg} \cdot 9.81\,\text{m/s}^2 \cdot 20.0\,\text{m}$

   $= 60.0\,\text{J} + 235.4\,\text{J}$

   $\approx 295\,\text{J}$

   Answer

   $295\,\text{J}$

   Three significant figures like the initial data.

   Reasonableness

   With rounded numbers

   $60\,\text{J} + 1.2\,\text{kg} \cdot 10\,\text{m/s}^2 \cdot 20\,\text{m}$

   $= 60\,\text{J} + 240\,\text{J}$

   $= 300\,\text{J}$

   The potential energy is greater than the kinetic energy, which suits a gull flying high but slowly.

4. Exercise 10-4

   A climber whose mass with equipment is $72\,\text{kg}$ climbs $12\,\text{m}$ up a climbing wall. By how much does his potential energy increase? Answer in kilojoules.

   kJCheck
   Answer

   $8.5\,\text{kJ}$

   Show the solution

   Given

   mass $m = 72\,\text{kg}$, change in height $\Delta h = 12\,\text{m}$ upward, $g = 9.81\,\text{m/s}^2$

   Find

   the change in potential energy $\Delta E_\text{p}$ in kilojoules

   Principle

   Potential energy is proportional to height, so its change depends only on the change in height, not on the zero level. The climber goes up, so the potential energy increases.

   Formula

   $E_\text{p} = m g h$, so $\Delta E_\text{p} = m g h_2 - m g h_1$, that is, $\Delta E_\text{p} = m g\,\Delta h$

   Conversions

   All quantities are in SI units, so the result comes in joules. Finally $1\,\text{kJ} = 1000\,\text{J}$.

   Substitution

   $\Delta E_\text{p}$$= 72\,\text{kg} \cdot 9.81\,\text{m/s}^2 \cdot 12\,\text{m}$

   $\approx 8476\,\text{J}$

   $\approx 8.48\,\text{kJ}$

   Answer

   $8.5\,\text{kJ}$

   Two significant figures like the mass $72\,\text{kg}$ and the height $12\,\text{m}$.

   Reasonableness

   The climber's weight is about $700\,\text{N}$, and $700\,\text{N} \cdot 12\,\text{m} \approx 8400\,\text{J}$. The result does not depend on the height from which the climber started, because only the change is asked.

5. Exercise 10-5

   A bucket with a mass of $4.0\,\text{kg}$ hangs in a well $6.5\,\text{m}$ below the ground. The zero level is the ground.

   How great is the potential energy of the bucket?

   JCheck
   Answer

   $-2.6 \cdot 10^{2}\,\text{J}$

   Show the solution

   Given

   mass $m = 4.0\,\text{kg}$, height $h = -6.5\,\text{m}$ (the bucket is $6.5\,\text{m}$ below the zero level, that is, the ground), $g = 9.81\,\text{m/s}^2$

   Find

   the potential energy $E_\text{p}$ of the bucket in joules

   Principle

   Potential energy is measured from the chosen zero level. Below the zero level the height is negative, so the potential energy is negative too: the bucket has less potential energy than at ground level.

   Formula

   $E_\text{p} = m g h$. The formula is already solved for the quantity asked.

   Conversions

   All quantities are in SI units, so no conversions are needed.

   Substitution

   $E_\text{p}$$= 4.0\,\text{kg} \cdot 9.81\,\text{m/s}^2 \cdot (-6.5\,\text{m})$

   $\approx -255\,\text{J}$

   Answer

   $-2.6 \cdot 10^{2}\,\text{J}$

   To two significant figures, like the mass $4.0\,\text{kg}$ and the depth $6.5\,\text{m}$.

   Reasonableness

   Lifting the bucket to ground level takes about $40\,\text{N} \cdot 6.5\,\text{m} = 260\,\text{J}$, and that is exactly how much lower the potential energy of the bucket is than its value zero at ground level. The sign is negative, as it should be below the zero level.

6. Exercise 10-6

   A car drives on a level road first at speed $v$ and then at speed $2v$. The mass of the car is the same in both cases.

   Which statement about the change in the kinetic energy of the car is correct?

   Choose an option.

   • The kinetic energy becomes four times as great, because it is proportional to the square of the speed.
   • The kinetic energy stays the same, because it depends only on the mass of the car.
   • The kinetic energy doubles, because it is proportional to the speed.
   • The kinetic energy becomes eight times as great, because it is proportional to the third power of the speed.

   Answer
   I have done my solution.
   Answer

   The kinetic energy becomes four times as great, because it is proportional to the square of the speed.

   Show the solution

   Find

   how the kinetic energy of the car changes when the speed doubles and the mass stays the same

   Principle

   Kinetic energy $E_\text{k} = \tfrac{1}{2}\,mv^2$ depends on the square of the speed. When the speed becomes twice as great, the square grows by the factor $2^2 = 4$, and the mass stays the same.

   Model answer

   The kinetic energy becomes four times as great, because $\tfrac{1}{2}\,m(2v)^2 = 4 \cdot \tfrac{1}{2}\,mv^2$. The statement about doubling is wrong, because kinetic energy is not proportional to the speed but to its square. The statement about becoming eight times as great is wrong, because in the formula the speed is raised to the second power, not the third. The statement that it stays the same is wrong, because kinetic energy depends on the speed as well as the mass.

7. Exercise 10-7

   A lift raises a person whose mass is $80\,\text{kg}$ to a higher floor. The potential energy of the person increases by $9.4\,\text{kJ}$.

   By how many metres does the lift raise the person?

   mCheck
   Answer

   $12\,\text{m}$

   Show the solution

   Given

   mass $m = 80\,\text{kg}$, change in potential energy $\Delta E_\text{p} = 9.4\,\text{kJ}$, $g = 9.81\,\text{m/s}^2$

   Find

   the lifting height $\Delta h$ in $\text{m}$

   Principle

   The change in potential energy is proportional to the change in height. When the change and the mass are known, the change in height is solved from the formula.

   Formula

   $\Delta E_\text{p} = m g\,\Delta h$, so

   $\Delta h$$= \dfrac{\Delta E_\text{p}}{m g}$

   Conversions

   $9.4\,\text{kJ}$$= 9.4 \cdot 1000\,\text{J}$

   $= 9400\,\text{J}$

   Substitution

   $\Delta h$$= \dfrac{9400\,\text{J}}{80\,\text{kg} \cdot 9.81\,\text{m/s}^2}$

   $= \dfrac{9400\,\text{J}}{784.8\,\text{N}}$

   $\approx 11.98\,\text{m}$

   Answer

   $12\,\text{m}$

   Two significant figures like the mass and the energy.

   Reasonableness

   With rounded numbers

   $80\,\text{kg} \cdot 10\,\text{m/s}^2 \cdot 12\,\text{m}$

   $= 9600\,\text{J}$

   $\approx 9.4\,\text{kJ}$

   About $12\,\text{m}$ corresponds to the height of four floors, which suits the distance travelled by a lift.

8. Exercise 10-8

   A bucket is $3.0\,\text{m}$ below the ground in a well and moves downward at a speed of $2.0\,\text{m/s}$. The zero level is the ground.

   Which statement about the kinetic energy $E_\text{k}$ and the potential energy $E_\text{p}$ of the bucket is correct?

   Choose an option.

   • $E_\text{k} > 0$ and $E_\text{p} < 0$, because kinetic energy does not depend on direction and the height is negative.
   • $E_\text{k} > 0$ and $E_\text{p} > 0$, because energy is always positive when an object moves.
   • $E_\text{k} < 0$ and $E_\text{p} < 0$, because the bucket moves down and is below the zero level.
   • $E_\text{k} < 0$ and $E_\text{p} > 0$, because the velocity points down and the depth is a positive number.

   Answer
   I have done my solution.
   Answer

   $E_\text{k} > 0$ and $E_\text{p} < 0$, because kinetic energy does not depend on direction and the height is negative.

   Show the solution

   Find

   the signs of the kinetic energy and the potential energy of the bucket when the bucket is below the zero level and moving

   Principle

   Kinetic energy $E_\text{k} = \tfrac{1}{2}\,mv^2$ contains the square of the speed, so it is never negative and does not depend on the direction of motion. Potential energy $E_\text{p} = m g h$ is calculated from the height above the zero level, and below the zero level $h < 0$.

   Model answer

   The kinetic energy is positive and the potential energy is negative. The bucket is moving, so $E_\text{k} > 0$, and it is below the zero level, so $E_\text{p} = m g h < 0$. The statement that both are negative is wrong, because downward motion does not make kinetic energy negative. The statement that both are positive is wrong, because below the zero level the potential energy is negative. The statement that the kinetic energy is negative and the potential energy positive is wrong, because it turns the signs the wrong way round: in the formula $h$ is the height from the zero level, not the depth.

Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.

1. Exercise 10-9Core

   The mass of a delivery van is $1350\,\text{kg}$, and it drives on a motorway at a speed of $90\,\text{km/h}$.

   How great is the kinetic energy of the van? Answer in kilojoules.

   kJCheck
   Answer

   $4.2 \cdot 10^{2}\,\text{kJ}$

   Show the solution

   Given

   mass $m = 1350\,\text{kg}$, speed $v = 90\,\text{km/h}$

   Find

   the kinetic energy $E_\text{k}$ of the van in kilojoules

   Principle

   Kinetic energy is proportional to the mass and the square of the speed. The joule is $\text{kg} \cdot \text{m}^2/\text{s}^2$, so the speed must first be converted to the unit $\text{m/s}$.

   Formula

   $E_\text{k} = \tfrac{1}{2}\,mv^2$. The formula is already solved for the quantity asked.

   Conversions

   $90\,\text{km/h}$$= \dfrac{90 \cdot 1000\,\text{m}}{3600\,\text{s}}$

   $= 25\,\text{m/s}$

   and finally $1\,\text{kJ} = 1000\,\text{J}$.

   Substitution

   $E_\text{k}$$= \tfrac{1}{2} \cdot 1350\,\text{kg} \cdot (25\,\text{m/s})^2$

   $= 675\,\text{kg} \cdot 625\,\text{m}^2/\text{s}^2$

   $= 421\,875\,\text{J}$

   $\approx 422\,\text{kJ}$

   Answer

   $4.2 \cdot 10^{2}\,\text{kJ}$

   To two significant figures, like the speed $90\,\text{km/h}$.

   Reasonableness

   With rounded numbers

   $\tfrac{1}{2} \cdot 1400\,\text{kg} \cdot 600\,\text{m}^2/\text{s}^2$

   $= 420\,\text{kJ}$

   If the speed were substituted in kilometres per hour, the result would be almost $13$ times too large, because $3.6^2 \approx 13$.

2. Exercise 10-10Core

   The mass of a tennis ball is $57\,\text{g}$. After a serve the kinetic energy of the ball is $82\,\text{J}$. What is the speed of the ball?

   m/sCheck
   Answer

   $54\,\text{m/s}$

   Show the solution

   Given

   mass $m = 57\,\text{g}$, kinetic energy $E_\text{k} = 82\,\text{J}$

   Find

   the speed $v$ of the ball in $\text{m/s}$

   Principle

   Kinetic energy depends on the mass and the speed. When the energy and the mass are known, the speed is solved from the expression for kinetic energy. The speed is positive, because we ask for its magnitude.

   Formula

   $E_\text{k} = \tfrac{1}{2}\,mv^2$, so $v^2 = \dfrac{2E_\text{k}}{m}$ and $v = \sqrt{\dfrac{2E_\text{k}}{m}}$

   Conversions

   $m$$= 57\,\text{g}$

   $= 57 : 1000\,\text{kg}$

   $= 0.057\,\text{kg}$

   Substitution

   $v$$= \sqrt{\dfrac{2 \cdot 82\,\text{J}}{0.057\,\text{kg}}}$

   $= \sqrt{2877\,\text{m}^2/\text{s}^2}$

   $\approx 53.6\,\text{m/s}$

   Answer

   $54\,\text{m/s}$

   Two significant figures like the mass $57\,\text{g}$ and the energy $82\,\text{J}$.

   Reasonableness

   $54\,\text{m/s} \approx 190\,\text{km/h}$, which is the speed of a hard serve by a top player. Calculated back,

   $\tfrac{1}{2} \cdot 0.057\,\text{kg} \cdot (54\,\text{m/s})^2$

   $\approx 83\,\text{J}$

   close to the given energy.

3. Exercise 10-11

   A book with a mass of $1.20\,\text{kg}$ is on a shelf $1.90\,\text{m}$ above the floor. In the room there is a table whose surface is $0.75\,\text{m}$ above the floor.

   How great is the potential energy of the book when the surface of the table is chosen as the zero level?

   JCheck
   Answer

   $13.5\,\text{J}$

   Show the solution

   Given

   mass $m = 1.20\,\text{kg}$, height of the shelf from the floor $1.90\,\text{m}$, height of the table surface from the floor $0.75\,\text{m}$, $g = 9.81\,\text{m/s}^2$; the zero level is the surface of the table

   Find

   the potential energy $E_\text{p}$ of the book relative to the surface of the table in joules

   Principle

   In potential energy the height is measured from the chosen zero level. The zero level is the surface of the table, so the height is the difference between the heights of the shelf and the table, not the height of the shelf from the floor.

   Formula

   $E_\text{p} = m g h$, where $h = h_\text{shelf} - h_\text{table}$. So $E_\text{p} = m g\,(h_\text{shelf} - h_\text{table})$

   Conversions

   All quantities are in SI units, so no conversions are needed.

   Substitution

   $E_\text{p}$$= 1.20\,\text{kg} \cdot 9.81\,\text{m/s}^2 \cdot (1.90 - 0.75)\,\text{m}$

   $= 1.20\,\text{kg} \cdot 9.81\,\text{m/s}^2 \cdot 1.15\,\text{m}$

   $\approx 13.5\,\text{J}$

   Answer

   $13.5\,\text{J}$

   Three significant figures like the mass and the difference in height $1.15\,\text{m}$.

   Reasonableness

   Relative to the floor the potential energy would be

   $1.20\,\text{kg} \cdot 9.81\,\text{m/s}^2 \cdot 1.90\,\text{m}$

   $\approx 22.4\,\text{J}$

   Relative to the table it is smaller, because the book is lower relative to the surface of the table than relative to the floor. The difference $8.8\,\text{J}$ is exactly the share of the height of the table, $0.75\,\text{m}$.

4. Exercise 10-12

   A remote-controlled model plane with a mass of $1.80\,\text{kg}$ flies $40.0\,\text{m}$ above the ground. Its mechanical energy relative to the ground is $905\,\text{J}$.

   What is the speed of the plane?

   m/sCheck
   Answer

   $14.9\,\text{m/s}$

   Show the solution

   Given

   mass $m = 1.80\,\text{kg}$, height $h = 40.0\,\text{m}$ above the ground, mechanical energy $E_\text{mech} = 905\,\text{J}$ relative to the ground, $g = 9.81\,\text{m/s}^2$

   Find

   the speed $v$ of the plane in $\text{m/s}$

   Principle

   Mechanical energy is the sum of kinetic and potential energy. When the potential energy is calculated from the height and subtracted from the mechanical energy, the kinetic energy is left, and the speed is solved from it.

   Formula

   $E_\text{mech} = \tfrac{1}{2}\,mv^2 + m g h$, so

   $\tfrac{1}{2}\,mv^2$$= E_\text{mech} - m g h$

   $v$$= \sqrt{\dfrac{2\,(E_\text{mech} - m g h)}{m}}$

   Conversions

   All quantities are in SI units, so no conversions are needed.

   Substitution

   $m g h$$= 1.80\,\text{kg} \cdot 9.81\,\text{m/s}^2 \cdot 40.0\,\text{m}$

   $\approx 706.3\,\text{J}$

   so

   $v$$= \sqrt{\dfrac{2 \cdot (905 - 706.3)\,\text{J}}{1.80\,\text{kg}}}$

   $= \sqrt{220.8\,\text{m}^2/\text{s}^2}$

   $\approx 14.9\,\text{m/s}$

   Answer

   $14.9\,\text{m/s}$

   Three significant figures like the initial data.

   Reasonableness

   Calculated back,

   $\tfrac{1}{2} \cdot 1.80\,\text{kg} \cdot (14.9\,\text{m/s})^2$

   $\approx 200\,\text{J}$

   $200\,\text{J} + 706\,\text{J} \approx 905\,\text{J}$. The speed $15\,\text{m/s} \approx 54\,\text{km/h}$ is normal for a model plane.

5. Exercise 10-13

   A rubber ball with a mass of $0.057\,\text{kg}$ was dropped from above the floor. With video analysis the height $h$ of the ball from the floor and its speed $v$ were measured as functions of the time $t$. Air resistance was so small that it is ignored. Example data is below.

   a) For each moment, calculate the kinetic energy, the potential energy and the mechanical energy. The zero level is the floor.

   b) Describe how the three energies change during the drop.

   c) Calculate the mean $E_{\text{mech,av}}$ of the mechanical energy over all moments.

   Write the mean of part c) in the answer box.

   Height $h$ and speed $v$ of the dropped ball as functions of the time $t$ (example data)

   $t\,(\text{s})$	$h\,(\text{m})$	$v\,(\text{m/s})$
   0.0	1.80	0.00
   0.1	1.75	0.98
   0.2	1.60	1.99
   0.3	1.36	2.91
   0.4	1.02	3.95
   0.5	0.58	4.88
   0.6	0.04	5.92

   JCheck
   Answer

   $1.0\,\text{J}$

   Show the solution

   Given

   mass $m = 0.057\,\text{kg}$, seven measurements $(t, h, v)$ of the table at the moments $0.0\,\text{s}$, $0.1\,\text{s}$, $\ldots$, $0.6\,\text{s}$; the zero level is the floor; $g = 9.81\,\text{m/s}^2$

   Find

   $E_\text{k}$, $E_\text{p}$ and $E_\text{mech}$ at each moment, how they change with time, and the mean $E_{\text{mech,av}}$ of the mechanical energy in joules

   Principle

   Kinetic energy is calculated from the speed and potential energy from the height above the floor. Mechanical energy is their sum. Air resistance is ignored, so the mechanical energy should stay the same, and measurement errors cause only a small scatter.

   Formula

   $E_\text{k} = \tfrac{1}{2}\,mv^2$, $E_\text{p} = m g h$ and $E_\text{mech} = E_\text{k} + E_\text{p}$. The mean is the sum of the mechanical energies divided by the number of measurements, $7$.

   Conversions

   All quantities are in SI units, so no conversions are needed.

   Substitution

   For example at the moment $t = 0.3\,\text{s}$:

   $$\begin{aligned} E_\text{k} &= \tfrac{1}{2} \cdot 0.057\,\text{kg} \cdot (2.91\,\text{m/s})^2 \\ &= 0.241\,\text{J} \\ E_\text{p} &= 0.057\,\text{kg} \cdot 9.81\,\text{m/s}^2 \cdot 1.36\,\text{m} \\ &= 0.760\,\text{J} \\ E_\text{mech} &= 0.241\,\text{J} + 0.760\,\text{J} \\ &= 1.001\,\text{J} \end{aligned}$$

   Similarly for the other moments (in joules):

   $$\begin{array}{c|ccc} t\,(\text{s}) & E_\text{k} & E_\text{p} & E_\text{mech} \\ \hline 0.0 & 0.000 & 1.007 & 1.007 \\ 0.1 & 0.027 & 0.979 & 1.006 \\ 0.2 & 0.113 & 0.895 & 1.008 \\ 0.3 & 0.241 & 0.760 & 1.001 \\ 0.4 & 0.445 & 0.570 & 1.015 \\ 0.5 & 0.679 & 0.324 & 1.003 \\ 0.6 & 0.999 & 0.022 & 1.021 \end{array}$$

   Part b): the kinetic energy grows from almost zero to about one joule, the potential energy decreases by the same amount, and their sum, the mechanical energy, stays almost the same. It varies by less than $1.5\,\%$ of the mean, which is of the order of the measurement error.
   Part c): the sum of the mechanical energies is $7.061\,\text{J}$, so

   $E_{\text{mech,av}}$$= 7.061\,\text{J} : 7$

   $\approx 1.009\,\text{J}$

   Answer

   $1.0\,\text{J}$

   Two significant figures like the mass $0.057\,\text{kg}$.

   Reasonableness

   At the start the ball is at rest at a height of $1.80\,\text{m}$, so

   $E_\text{mech}$$= m g h$

   $= 0.057\,\text{kg} \cdot 9.81\,\text{m/s}^2 \cdot 1.80\,\text{m}$

   $\approx 1.01\,\text{J}$

   the same as the mean. At the last moment the ball is almost at the floor, and almost all the energy is kinetic energy.

6. Exercise 10-14

   A diver whose mass with equipment is $70\,\text{kg}$ has jumped into the water from a tower. When he is $1.5\,\text{m}$ below the surface of the water, his speed is $4.0\,\text{m/s}$. The zero level is the water surface. The task calculates only the mechanical energy $E_\text{k} + E_\text{p}$ as defined.

   A student calculated the mechanical energy of the diver like this: “$E_\text{mech} = \tfrac{1}{2}\,mv^2 + mgh = \tfrac{1}{2} \cdot 70\,\text{kg} \cdot (4.0\,\text{m/s})^2 + 70\,\text{kg} \cdot 9.81\,\text{m/s}^2 \cdot 1.5\,\text{m} \approx 1590\,\text{J}$.”

   a) Explain the student’s error.

   b) Calculate the mechanical energy of the diver correctly.

   Write the result of part b) in the answer box.

   JCheck
   Answer

   $-4.7 \cdot 10^{2}\,\text{J}$

   Show the solution

   Given

   mass $m = 70\,\text{kg}$, depth $1.5\,\text{m}$ below the zero level, that is, the water surface, speed $v = 4.0\,\text{m/s}$, $g = 9.81\,\text{m/s}^2$

   Find

   the student’s error and the mechanical energy $E_\text{mech}$ of the diver in joules

   Principle

   Potential energy is calculated from the height above the zero level. Below the water surface the height is negative, $h = -1.5\,\text{m}$, so the potential energy is negative. The student substituted the depth into the formula as a positive number and got a positive potential energy. Kinetic energy is positive whatever the direction.

   Formula

   $E_\text{mech}$$= E_\text{k} + E_\text{p}$

   $= \tfrac{1}{2}\,mv^2 + m g h$

   where $h = -1.5\,\text{m}$

   Conversions

   All quantities are in SI units, so no conversions are needed.

   Substitution

   The height is $h = -1.5\,\text{m}$, so

   $$\begin{aligned} E_\text{k} &= \tfrac{1}{2} \cdot 70\,\text{kg} \cdot (4.0\,\text{m/s})^2 \\ &= 560\,\text{J} \\ E_\text{p} &= 70\,\text{kg} \cdot 9.81\,\text{m/s}^2 \cdot (-1.5\,\text{m}) \\ &\approx -1030\,\text{J} \\ E_\text{mech} &= 560\,\text{J} - 1030\,\text{J} \\ &= -470\,\text{J} \end{aligned}$$

   Answer

   $-4.7 \cdot 10^{2}\,\text{J}$

   With two significant figures like the mass and the speed. The error: the student used the depth as a positive height, although $h < 0$ below the zero level.

   Reasonableness

   The diver is below the zero level, so the potential energy is negative, and its absolute value $1030\,\text{J}$ is greater than the kinetic energy $560\,\text{J}$. That is why the sum is also negative. The student’s positive result $1590\,\text{J}$ cannot be right, because below the zero level the potential energy decreases the mechanical energy.

7. Exercise 10-15

   The kinetic energy of a cart was measured at different speeds. The mass of the cart was not written down, so it has to be determined from the measurement data. Example data is below.

   Draw the $(v^2, E_\text{k})$ graph and fit to it a line through the origin. Using the formula $E_\text{k} = \tfrac{1}{2}\,mv^2$, justify why the slope of the line gives the mass of the cart, and determine the mass.

   Write the mass of the cart in the answer box.

   Speed $v$ and kinetic energy $E_\text{k}$ of the cart (example data)

   $v\,(\text{m/s})$	$E_\text{k}\,(\text{J})$
   0.40	0.043
   0.80	0.17
   1.20	0.37
   1.60	0.66
   2.00	1.04

   kgCheck
   Answer

   $0.52\,\text{kg}$

   Show the solution

   Given

   five measurements $(v, E_\text{k})$ of the example data: $(0.40\,\text{m/s}$; $0.043\,\text{J})$, $(0.80\,\text{m/s}$; $0.17\,\text{J})$, $(1.20\,\text{m/s}$; $0.37\,\text{J})$, $(1.60\,\text{m/s}$; $0.66\,\text{J})$ and $(2.00\,\text{m/s}$; $1.04\,\text{J})$

   Find

   the mass $m$ of the cart in $\text{kg}$

   Principle

   Kinetic energy is $E_\text{k} = \tfrac{1}{2}\,mv^2$. When $E_\text{k}$ is plotted as a function of the square of the speed $v^2$, the graph is a line through the origin whose slope is $\tfrac{1}{2}\,m$. The mass is therefore twice the slope.

   Reading the graph

   First calculate $v^2$: $0.16$, $0.64$, $1.44$, $2.56$ and $4.00\,\text{m}^2/\text{s}^2$. Plot the points $(v^2, E_\text{k})$ and fit to them a line through the origin. A distant point is read from the line: when $v^2 = 4.00\,\text{m}^2/\text{s}^2$, the value of the line is about $E_\text{k} = 1.04\,\text{J}$.

   Formula

   slope

   $k$$= \dfrac{\Delta E_\text{k}}{\Delta (v^2)}$

   $= \tfrac{1}{2}\,m$

   so $m = 2k$

   Conversions

   All quantities are in SI units, so no conversions are needed. The unit of the slope is $\text{J} : (\text{m}^2/\text{s}^2) = \text{kg}$.

   Substitution

   $k$$= \dfrac{1.04\,\text{J} - 0\,\text{J}}{4.00\,\text{m}^2/\text{s}^2 - 0\,\text{m}^2/\text{s}^2}$

   $= 0.26\,\text{kg}$

   so

   $m$$= 2 \cdot 0.26\,\text{kg}$

   $= 0.52\,\text{kg}$

   Answer

   $0.52\,\text{kg}$

   Two significant figures like the measured values.

   Reasonableness

   Check with one point:

   $\tfrac{1}{2} \cdot 0.52\,\text{kg} \cdot (1.20\,\text{m/s})^2$

   $= 0.37\,\text{J}$

   the same as in the table. A mass of $0.52\,\text{kg}$ is normal for a laboratory cart.

Optional: application, two ideas together or a new situation where you reason for yourself.

1. Exercise 10-16

   A lorry with a mass of $12.0\,\text{t}$ drives in a built-up area at a speed of $40.0\,\text{km/h}$.

   At what speed should a passenger car with a mass of $1.20\,\text{t}$ drive for its kinetic energy to be equal to that of the lorry? Answer in km/h.

   km/hCheck
   Answer

   $126\,\text{km/h}$

   Show the solution

   Given

   mass of the lorry $m_2 = 12.0\,\text{t}$ and speed $v_2 = 40.0\,\text{km/h}$, mass of the passenger car $m_1 = 1.20\,\text{t}$

   Find

   the speed $v_1$ of the passenger car in $\text{km/h}$ when the kinetic energies are equal

   Principle

   Kinetic energy is proportional to the mass and the square of the speed. The lighter car has to drive faster, but because the speed is squared, ten times the mass is made up for by only $\sqrt{10}$ times the speed.

   Formula

   $\tfrac{1}{2}\,m_1 v_1^2 = \tfrac{1}{2}\,m_2 v_2^2$, so $v_1^2 = \dfrac{m_2}{m_1}\,v_2^2$ and $v_1 = v_2\sqrt{\dfrac{m_2}{m_1}}$

   Conversions

   The formula has only the ratio of the masses and the speed is multiplied by it, so the tonnes need not be converted to kilograms nor the speed to $\text{m/s}$: the units are preserved.

   Substitution

   $v_1$$= 40.0\,\text{km/h} \cdot \sqrt{\dfrac{12.0\,\text{t}}{1.20\,\text{t}}}$

   $= 40.0\,\text{km/h} \cdot \sqrt{10.0}$

   $\approx 126.5\,\text{km/h}$

   Answer

   $126\,\text{km/h}$

   Three significant figures like the initial data.

   Reasonableness

   Check in joules: $40.0\,\text{km/h} = 11.11\,\text{m/s}$ and $126.5\,\text{km/h} = 35.14\,\text{m/s}$, so

   $\tfrac{1}{2} \cdot 12\,000\,\text{kg} \cdot (11.11\,\text{m/s})^2$

   $\approx 741\,\text{kJ}$

   $\tfrac{1}{2} \cdot 1200\,\text{kg} \cdot (35.14\,\text{m/s})^2$

   $\approx 741\,\text{kJ}$

   A lorry at city speed thus carries as much kinetic energy as a passenger car well over the motorway speed.

2. Exercise 10-17

   The bob of a pendulum, with a mass of $0.500\,\text{kg}$, hangs at the end of a string $1.20\,\text{m}$ long. The bob is pulled aside so that the taut string is at an angle of $40^\circ$ to the vertical.

   How great is the potential energy of the bob when its lowest point is chosen as the zero level?

   JCheck
   Answer

   $1.38\,\text{J}$

   Show the solution

   Given

   mass $m = 0.500\,\text{kg}$, length of the string $L = 1.20\,\text{m}$, angle of the string to the vertical $\alpha = 40^\circ$, $g = 9.81\,\text{m/s}^2$; the zero level is the lowest point of the bob

   Find

   the potential energy $E_\text{p}$ of the bob in joules

   Principle

   For potential energy we need the height of the bob above its lowest point. The bob moves along a circular arc whose radius is the length of the string. Measured from the point of suspension, the bob is at its lowest a distance $L$ below it, and when pulled aside a distance $L\cos\alpha$ below it, so the height is the difference of these.

   Formula

   $h$$= L - L\cos\alpha$

   $= L\,(1 - \cos\alpha)$

   $E_\text{p} = m g h$, so $E_\text{p} = m g L\,(1 - \cos\alpha)$

   Conversions

   All quantities are in SI units, so no conversions are needed. The calculator must be in degrees.

   Substitution

   $h$$= 1.20\,\text{m} \cdot (1 - \cos 40^\circ)$

   $\approx 1.20\,\text{m} \cdot 0.2340$

   $\approx 0.2808\,\text{m}$

   so

   $E_\text{p}$$= 0.500\,\text{kg} \cdot 9.81\,\text{m/s}^2 \cdot 0.2808\,\text{m}$

   $\approx 1.377\,\text{J}$

   Answer

   $1.38\,\text{J}$

   Three significant figures like the mass $0.500\,\text{kg}$ and the length of the string $1.20\,\text{m}$.

   Reasonableness

   The height $0.28\,\text{m}$ is clearly smaller than the length of the string, as it should be when the angle is under $90^\circ$; at the angle $90^\circ$ the height would be the whole $1.20\,\text{m}$. With rounded numbers $5\,\text{N} \cdot 0.28\,\text{m} \approx 1.4\,\text{J}$.

3. Exercise 10-18

   A roller skater with a mass of $55.0\,\text{kg}$ arrives at the foot of a ramp at a speed of $7.00\,\text{m/s}$. She does not push for extra speed, and she stops for a moment at a height of $2.10\,\text{m}$ above the foot of the ramp.

   How much mechanical energy changed into other energy, such as internal energy, during the rise?

   JCheck
   Answer

   $214\,\text{J}$

   Show the solution

   Given

   mass $m = 55.0\,\text{kg}$, speed at the foot of the ramp $v = 7.00\,\text{m/s}$, height at the highest point $h = 2.10\,\text{m}$, where the speed is zero, $g = 9.81\,\text{m/s}^2$; the zero level is the foot of the ramp

   Find

   the mechanical energy that changed into other energy, in joules

   Principle

   Without friction and air resistance the mechanical energy would stay the same. Friction and air resistance are not conservative, and they change part of the mechanical energy into internal energy. The amount changed is the decrease in mechanical energy from the foot to the highest point. At the foot there is only kinetic energy ($h = 0$), at the highest point only potential energy ($v = 0$).

   Formula

   $E_\text{other} = E_\text{mech,1} - E_\text{mech,2}$, where $E_\text{mech,1} = \tfrac{1}{2}\,mv^2$ and $E_\text{mech,2} = m g h$. So $E_\text{other} = \tfrac{1}{2}\,mv^2 - m g h$

   Conversions

   All quantities are in SI units, so no conversions are needed.

   Substitution

   $\tfrac{1}{2}\,m v^2$$= \tfrac{1}{2} \cdot 55.0\,\text{kg} \cdot (7.00\,\text{m/s})^2$

   $= 1347.5\,\text{J}$

   $m g h$$= 55.0\,\text{kg} \cdot 9.81\,\text{m/s}^2 \cdot 2.10\,\text{m}$

   $\approx 1133.1\,\text{J}$

   so

   $E_\text{other}$$= 1347.5\,\text{J} - 1133.1\,\text{J}$

   $\approx 214\,\text{J}$

   Answer

   $214\,\text{J}$

   Three significant figures like the initial data.

   Reasonableness

   Without friction the skater would rise to the height

   $\tfrac{v^2}{2g}$$= \tfrac{49\,\text{m}^2/\text{s}^2}{19.62\,\text{m/s}^2}$

   $\approx 2.50\,\text{m}$

   She rises $2.10\,\text{m}$, about $16\,\%$ less, and the changed energy $214\,\text{J}$ is about $16\,\%$ of the original $1347.5\,\text{J}$. The result is positive, because friction only decreases the mechanical energy.

4. Exercise 10-19

   A tennis ball with a mass of $0.058\,\text{kg}$ was dropped onto a hard floor. From a video, the highest point of the ball above the floor was measured before the drop and after each bounce. The ball moves vertically, and at the highest point its speed is zero. Example data is below.

   a) Calculate the mechanical energy of the ball at each highest point when the zero level is the floor.

   b) On average, what percentage of the mechanical energy changes into other energy in one bounce?

   c) How much mechanical energy has changed into other energy during the first four bounces? Where has the energy gone?

   d) Another student chooses as the zero level the surface of a table that is $1.00\,\text{m}$ above the floor. With her zero level, calculate the mechanical energy of the ball at the moment of the drop and after the fourth bounce. Justify why the result of part c) is still the same.

   Write the energy of part c) in the answer box.

   Highest point $h$ of the ball above the floor (example data)

   Bounces	$h\,(\text{m})$
   0	1.50
   1	1.09
   2	0.80
   3	0.57
   4	0.42

   JCheck
   Answer

   $0.61\,\text{J}$

   Show the solution

   Given

   mass $m = 0.058\,\text{kg}$, highest points $1.50\,\text{m}$, $1.09\,\text{m}$, $0.80\,\text{m}$, $0.57\,\text{m}$ and $0.42\,\text{m}$ above the floor; at the highest point $v = 0$; $g = 9.81\,\text{m/s}^2$

   Find

   the mechanical energy at the highest points, the average percentage, the energy $\Delta E$ changed in the first four bounces in joules, and why the results do not depend on the zero level

   Principle

   At the highest point the speed is zero, so the mechanical energy is only potential energy, $E_\text{mech} = E_\text{p} = m g h$. If mechanical energy were conserved, the ball would always return to the same height. The height decreases, so part of the mechanical energy changes into other energy: in the collision the ball and the floor warm up slightly, the ball vibrates and sound is produced. The amount changed is the decrease in mechanical energy.

   Formula

   $E_\text{mech} = m g h$. The energy changed is

   $\Delta E$$= E_\text{mech,0} - E_\text{mech,4}$

   $= m g\,(h_0 - h_4)$

   The share in one bounce is obtained from the ratio of the heights, because $E_\text{mech}$ is proportional to the height.

   Conversions

   All quantities are in SI units, so no conversions are needed.

   Substitution

   Part a):

   $m g$$= 0.058\,\text{kg} \cdot 9.81\,\text{m/s}^2$

   $= 0.569\,\text{N}$

   and $E_n = m g h_n$:

   $$\begin{aligned} E_0 &= 0.569\,\text{N} \cdot 1.50\,\text{m} \\ &= 0.853\,\text{J} \\ E_1 &= 0.569\,\text{N} \cdot 1.09\,\text{m} \\ &= 0.620\,\text{J} \\ E_2 &= 0.569\,\text{N} \cdot 0.80\,\text{m} \\ &= 0.455\,\text{J} \\ E_3 &= 0.569\,\text{N} \cdot 0.57\,\text{m} \\ &= 0.324\,\text{J} \\ E_4 &= 0.569\,\text{N} \cdot 0.42\,\text{m} \\ &= 0.239\,\text{J} \end{aligned}$$

   Part b): the ratios of consecutive heights are $\dfrac{1.09}{1.50} = 0.727$, $\dfrac{0.80}{1.09} = 0.734$, $\dfrac{0.57}{0.80} = 0.713$ and $\dfrac{0.42}{0.57} = 0.737$. The sum of the ratios is $2.911$, so the mean is $2.911 : 4 = 0.728$. In each bounce about $73\,\%$ of the mechanical energy is conserved and about $27\,\%$ changes into other energy.
   Part c):

   $$\begin{aligned} \Delta E &= 0.853\,\text{J} - 0.239\,\text{J} \\ &= 0.614\,\text{J} \end{aligned}$$

   Part d): the zero level is $1.00\,\text{m}$ above the floor, so the heights are $h' = h - 1.00\,\text{m}$:

   $$\begin{aligned} E_0' &= 0.569\,\text{N} \cdot 0.50\,\text{m} \\ &= 0.285\,\text{J} \\ E_4' &= 0.569\,\text{N} \cdot (-0.58\,\text{m}) \\ &= -0.330\,\text{J} \\ E_4' - E_0' &= -0.615\,\text{J} \end{aligned}$$

   The change is the same as with the floor as the zero level, because $m g\,(h_4' - h_0') = m g\,(h_4 - h_0)$: the height of the zero level cancels out of the difference. The value of the mechanical energy depends on the zero level, but its change does not.

   Answer

   $0.61\,\text{J}$

   Two significant figures like the mass $0.058\,\text{kg}$. The energy has gone into the internal energy of the ball, the floor and the air, and into sound.

   Reasonableness

   The total change is $1 - \dfrac{0.42}{1.50} = 72\,\%$ of the original energy, and $0.72 \cdot 0.853\,\text{J} = 0.614\,\text{J}$. Four bounces that each conserve $73\,\%$ give $0.73^4 \approx 0.28$, that is, $28\,\%$ remains, which matches the ratio of the table, $0.42 : 1.50 = 0.28$.

5. Exercise 10-20

   Along a sledding hill there were five measuring points where the height $h$ of the sledder above the bottom of the slope and the speed $v$ were measured. The total mass of the sledder and the sled is $48\,\text{kg}$. Example data is below.

   a) Calculate the mechanical energy at each measuring point when the zero level is the bottom of the slope.

   b) How much mechanical energy has changed into other energy between points 1 and 5?

   c) Between which points does the most mechanical energy change? Justify with an observation about the speed, and explain why the energy does not disappear anywhere.

   d) Choose the height of point 1 as the zero level. Calculate the mechanical energy at points 1 and 5 and justify why the result of part b) does not change.

   Write the energy of part b) in kilojoules in the answer box.

   Height $h$ and speed $v$ of the sledder at the measuring points (example data)

   Point	$h\,(\text{m})$	$v\,(\text{m/s})$
   1	18.0	1.5
   2	14.0	8.3
   3	10.0	11.5
   4	6.0	14.0
   5	2.0	15.8

   kJCheck
   Answer

   $1.6\,\text{kJ}$

   Show the solution

   Given

   mass $m = 48\,\text{kg}$, measuring points $(h, v)$: $(18.0\,\text{m}$; $1.5\,\text{m/s})$, $(14.0\,\text{m}$; $8.3\,\text{m/s})$, $(10.0\,\text{m}$; $11.5\,\text{m/s})$, $(6.0\,\text{m}$; $14.0\,\text{m/s})$ and $(2.0\,\text{m}$; $15.8\,\text{m/s})$; the zero level is the bottom of the slope; $g = 9.81\,\text{m/s}^2$

   Find

   the mechanical energy $E_\text{mech}$ at points 1–5, the energy $\Delta E$ changed into other energy between points 1 and 5 in kilojoules, and why the result does not depend on the zero level

   Principle

   At each point the mechanical energy is the sum of the kinetic energy and the potential energy. Without friction and air resistance it would stay the same, so its decrease tells how much energy has changed into other energy, that is, internal energy, because of friction and air resistance. The height is the vertical distance from the zero level, not the distance along the slope.

   Formula

   $E_\text{mech} = \tfrac{1}{2}\,mv^2 + m g h$ and $\Delta E = E_\text{mech,1} - E_\text{mech,5}$

   Conversions

   All quantities are in SI units, so the energies come in joules. Finally $1\,\text{kJ} = 1000\,\text{J}$.

   Substitution

   Part a), taking point 1 as an example, where

   $m g$$= 48\,\text{kg} \cdot 9.81\,\text{m/s}^2$

   $= 470.9\,\text{N}$

   :

   $$\begin{aligned} E_\text{k} &= \tfrac{1}{2} \cdot 48\,\text{kg} \cdot (1.5\,\text{m/s})^2 \\ &= 54\,\text{J} \\ E_\text{p} &= 470.9\,\text{N} \cdot 18.0\,\text{m} \\ &\approx 8476\,\text{J} \\ E_\text{mech} &\approx 8530\,\text{J} \end{aligned}$$

   All the points (in joules):

   $$\begin{array}{c|ccc} \text{Point} & E_\text{k} & E_\text{p} & E_\text{mech} \\ \hline 1 & 54 & 8476 & 8530 \\ 2 & 1653 & 6592 & 8246 \\ 3 & 3174 & 4709 & 7883 \\ 4 & 4704 & 2825 & 7529 \\ 5 & 5991 & 942 & 6933 \end{array}$$

   Part b):

   $$\begin{aligned} \Delta E &= 8530\,\text{J} - 6933\,\text{J} \\ &= 1597\,\text{J} \\ &\approx 1.6\,\text{kJ} \end{aligned}$$

   Part c): the differences in mechanical energy between consecutive points are $284\,\text{J}$, $363\,\text{J}$, $354\,\text{J}$ and $596\,\text{J}$. The most energy changes between points 4 and 5, where the speed is greatest, and the least between points 1 and 2, where the speed is smallest. This fits air resistance, which grows as the speed grows, and the friction of the snow acts in addition. The energy does not disappear: friction and air resistance warm the sled, the snow and the air, so the mechanical energy changes into their internal energy.
   Part d): the zero level is at the height of point 1, so the height of point 5 is

   $h'$$= 2.0\,\text{m} - 18.0\,\text{m}$

   $= -16.0\,\text{m}$

   and the height of point 1 is $0$:

   $$\begin{aligned} E_1' &= 54\,\text{J} + 0\,\text{J} \\ &= 54\,\text{J} \\ E_5' &= 5991\,\text{J} - 7534\,\text{J} \\ &= -1543\,\text{J} \\ E_1' - E_5' &= 1597\,\text{J} \end{aligned}$$

   The difference is the same, because the choice of the zero level adds the same number to both mechanical energies, and it cancels out of the difference.

   Answer

   $1.6\,\text{kJ}$

   Two significant figures like the mass $48\,\text{kg}$.

   Reasonableness

   If the mechanical energy were conserved, the speed at point 5 would be about $17.8\,\text{m/s}$, and the measured $15.8\,\text{m/s}$ is smaller than that. The energy changed is about $19\,\%$ of the original $8530\,\text{J}$, which is a reasonable share for a sledding hill.