Work and energy

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

1. Exercise 2-1Core

   A moped is pushed in the direction of motion with an $80\,\text{N}$ force over a distance of 25 metres. How much work does the pushing force do?

   kJCheck
   Answer

   $2.0\,\text{kJ}$

   Show the solution

   Given

   $F = 80\,\text{N}$ (in the direction of motion), $s = 25\,\text{m}$

   Find

   the work $W$ done by the force in kilojoules

   Principle

   The pushing force is in the direction of motion and constant, so it does work equal to the product of force and displacement. If the force were at an angle or changed along the way, another method would be needed.

   Formula

   $W = F s$, so the work is already solved for: $W = F s$

   Conversions

   The force is in newtons and the distance in metres, so the product is directly in joules ($1\,\text{J} = 1\,\text{N}\cdot\text{m}$). Finally $1\,\text{kJ} = 1000\,\text{J}$.

   Substitution

   $W$$= 80\,\text{N} \cdot 25\,\text{m}$

   $= 2000\,\text{J}$

   $= 2.000\,\text{kJ}$

   Answer

   $2.0\,\text{kJ}$

   Two significant figures like the given values $80\,\text{N}$ and $25\,\text{m}$.

   Reasonableness

   Rough estimate: $80\,\text{N} \approx 8\,\text{kg}$ in weight, and

   $8\,\text{kg}\cdot 10\,\tfrac{\text{m}}{\text{s}^2}\cdot 25\,\text{m}$

   $= 2000\,\text{J}$

   The same work would lift a $20\,\text{kg}$ object about $10\,\text{m}$:

   $20\,\text{kg}\cdot 9.81\,\tfrac{\text{m}}{\text{s}^2}\cdot 10\,\text{m}$

   $\approx 2.0\,\text{kJ}$

   $W$ is positive because the force is in the direction of motion.

2. Exercise 2-2Core

   A bookcase weighs $150\,\text{N}$. The bookcase is lifted slowly from the floor to a height of $1.2\,\text{m}$. How much work does the lifting force do?

   JCheck
   Answer

   $180\,\text{J}$

   Show the solution

   Given

   weight $G = 150\,\text{N}$, lifting height $h = 1.2\,\text{m}$, slow lift

   Find

   the work $W$ done by the lifting force in joules

   Principle

   In a slow lift the lifting force is equal to the weight and in the direction of motion, so the force is constant over the whole distance. The work is then the product of force and lifting height.

   Formula

   $W = F h$ and $F = G$, so $W = G h$

   Conversions

   The weight is in newtons and the height in metres, so no conversions are needed: $\text{N}\cdot\text{m} = \text{J}$.

   Substitution

   $W$$= 150\,\text{N} \cdot 1.2\,\text{m}$

   $= 180\,\text{J}$

   Answer

   $180\,\text{J}$

   Two significant figures like the height $1.2\,\text{m}$.

   Reasonableness

   Rough estimate: $150\,\text{N} \cdot 1\,\text{m} = 150\,\text{J}$, and the distance is a little over a metre. A weight of $150\,\text{N}$ corresponds to a mass of about $\tfrac{150\,\text{N}}{9.81\,\text{m/s}^2} \approx 15\,\text{kg}$, so the work corresponds to lifting a $15\,\text{kg}$ bookcase a little over a metre. $W$ is positive because the lifting force and the motion are both upwards.

3. Exercise 2-3

   The motor of a lift does $36\,\text{kJ}$ of work in 12 seconds. What is the power of the motor?

   kWCheck
   Answer

   $3.0\,\text{kW}$

   Show the solution

   Given

   work $W = 36\,\text{kJ}$, time $t = 12\,\text{s}$

   Find

   the power $P$ of the motor in kilowatts

   Principle

   Power tells you how fast work is done, that is, work per unit time. The motor does the work steadily in the given time, so the average power is the total work divided by the time.

   Formula

   $P = \dfrac{W}{t}$, so the power is already solved for: $P = \dfrac{W}{t}$

   Conversions

   $36\,\text{kJ}$$= 36 \cdot 1000\,\text{J}$

   $= 36\,000\,\text{J}$

   so that the result comes out in watts ($1\,\text{W} = 1\,\text{J/s}$). Finally $1\,\text{kW} = 1000\,\text{W}$.

   Substitution

   $P$$= \dfrac{36\,000\,\text{J}}{12\,\text{s}}$

   $= 3000\,\text{W}$

   $= 3.000\,\text{kW}$

   Answer

   $3.0\,\text{kW}$

   Two significant figures like the time $12\,\text{s}$.

   Reasonableness

   The units kilojoule and second give kilowatts directly:

   $\tfrac{36\,\text{kJ}}{12\,\text{s}}$$= 3\,\tfrac{\text{kJ}}{\text{s}}$

   $= 3\,\text{kW}$

   With this power you could lift a load of about $300\,\text{kg}$ at a steady one metre per second:

   $300\,\text{kg}\cdot 9.81\,\tfrac{\text{m}}{\text{s}^2}\cdot 1\,\tfrac{\text{m}}{\text{s}}$

   $\approx 2.9\,\text{kW}$

   This suits the motor of a lift.

4. Exercise 2-4

   The graph shows the magnitude of the force pulling a sledge, in the direction of motion, as a function of the distance travelled. Calculate the total work done by the force as the area under the graph.

   JCheck
   Answer

   $190\,\text{J}$

   Show the solution

   Given

   read from the graph, force $F_1 = 25\,\text{N}$ over $0\ldots 6\,\text{m}$ and $F_2 = 10\,\text{N}$ over $6\ldots 10\,\text{m}$, the force is in the direction of motion

   Find

   the total work $W$ done by the force in joules

   Principle

   The work is the area between the $(s,F)$ graph and the $s$-axis. The force is not constant over the whole distance, so the formula $W = F s$ does not apply directly to the whole distance. The area splits into two rectangles in which the force is constant.

   Reading the graph

   The first horizontal segment is on the grid line $F_1 = 25\,\text{N}$ and extends from $s = 0$ to $s = 6\,\text{m}$, where it drops vertically. The second segment is on the line $F_2 = 10\,\text{N}$ and extends from $s = 6\,\text{m}$ to $s = 10\,\text{m}$. The width of the second rectangle is

   $s_2$$= 10\,\text{m} - 6\,\text{m}$

   $= 4\,\text{m}$

   Formula

   $W = W_1 + W_2$, so $W = F_1 s_1 + F_2 s_2$

   Conversions

   The forces are in newtons and the distances in metres, so no conversions are needed: $\text{N}\cdot\text{m} = \text{J}$.

   Substitution

   $W$$= 25\,\text{N} \cdot 6\,\text{m} + 10\,\text{N} \cdot 4\,\text{m}$

   $= 150\,\text{J} + 40\,\text{J}$

   $= 190\,\text{J}$

   Answer

   $190\,\text{J}$

   Two significant figures, because the values of the graph were read from the grid lines as whole numbers.

   Reasonableness

   The average force over the whole $10\,\text{m}$ is

   $\dfrac{190\,\text{J}}{10\,\text{m}}$$= 19\,\text{N}$

   which lies between the smallest ($10\,\text{N}$) and the largest ($25\,\text{N}$) value of the force. If you calculated $25\,\text{N} \cdot 10\,\text{m} = 250\,\text{J}$ for the whole distance, you would get too large a result, because the force drops partway through.

5. Exercise 2-5

   A skier pulls a sled at a steady speed of $3.0\,\text{m/s}$ with a $20\,\text{N}$ force in the direction of motion. What is the power of the pulling force?

   WCheck
   Answer

   $60\,\text{W}$

   Show the solution

   Given

   force $F = 20\,\text{N}$ (in the direction of motion), speed $v = 3.0\,\text{m/s}$, steady motion

   Find

   the power $P$ of the pulling force in watts

   Principle

   Power is work per unit time. The motion is steady and the force is constant and in the direction of motion, so the distance travelled in one second is $v$ and the work done in one second is $F v$.

   Formula

   $P = \dfrac{W}{t}$ and $W = F s$, $s = v t$, so $P = \dfrac{F v t}{t} = F v$

   Conversions

   The force is in newtons and the speed in metres per second, so the product comes directly in watts: $1\,\text{N}\cdot\tfrac{\text{m}}{\text{s}} = 1\,\text{W}$.

   Substitution

   $P$$= 20\,\text{N} \cdot 3.0\,\tfrac{\text{m}}{\text{s}}$

   $= 60\,\text{W}$

   Answer

   $60\,\text{W}$

   Two significant figures like the speed $3.0\,\text{m/s}$.

   Reasonableness

   Check through time: in one second the sled travels $3.0\,\text{m}$, so the work is $20\,\text{N} \cdot 3.0\,\text{m} = 60\,\text{J}$ and the power is $\tfrac{60\,\text{J}}{1\,\text{s}} = 60\,\text{W}$. A power of a few tens of watts is typical for the pulling power at a brisk walking pace.

6. Exercise 2-6

   A student carries a backpack along a horizontal corridor at a steady speed. The force supporting the weight of the backpack points upwards.

   Why does this force do no work?

   Choose an option.

   • The backpack moves at a steady speed, so no force does work on it.
   • Work is done only when lifting, so horizontal motion requires no work.
   • The supporting force is equal to the weight, so it does no work.
   • The force points up, but the motion is sideways, so the force is not in the direction of motion.

   Answer
   I have done my solution.
   Answer

   The force points up, but the motion is sideways, so the force is not in the direction of motion.

   Show the solution

   Find

   why the supporting force does no work when a backpack is carried horizontally

   Principle

   The work $W = F s$ done by a force requires the force to be in the direction of motion. The force supporting the backpack points up, but the backpack moves sideways, so the force is perpendicular to the motion.

   Model answer

   The force points up, but the motion is sideways, so the force is not in the direction of motion and does no work. A steady speed does not explain it, because a force in the direction of motion does work even in steady motion, like the force pulling a sled. The magnitude of the force does not explain it either: in a slow lift a force equal to the weight does the work $W = G h$. Work is not limited to lifting, because a horizontal force pushing a moped does work.

7. Exercise 2-7

   A motor driving a conveyor belt pulls the belt at a steady speed with a $1.2\,\text{kN}$ force in the direction of motion. The power of the motor is $1.8\,\text{kW}$.

   At what speed does the belt move?

   m/sCheck
   Answer

   $1.5\,\text{m/s}$

   Show the solution

   Given

   force $F = 1.2\,\text{kN}$ (in the direction of motion), power $P = 1.8\,\text{kW}$, steady speed

   Find

   the speed $v$ of the belt in metres per second

   Principle

   The speed is steady and the force is constant and in the direction of motion, so the power is the product of force and velocity. The velocity can be solved when the power and the force are known.

   Formula

   $P = F v$, so $v = \dfrac{P}{F}$

   Conversions

   $1.8\,\text{kW} = 1800\,\text{W}$ and $1.2\,\text{kN} = 1200\,\text{N}$, so that the result comes out in the unit $\tfrac{\text{W}}{\text{N}} = \tfrac{\text{m}}{\text{s}}$.

   Substitution

   $v$$= \dfrac{1800\,\text{W}}{1200\,\text{N}}$

   $= 1.5\,\tfrac{\text{m}}{\text{s}}$

   Answer

   $1.5\,\text{m/s}$

   Two significant figures like the given values $1.2\,\text{kN}$ and $1.8\,\text{kW}$.

   Reasonableness

   Check in reverse: $1200\,\text{N} \cdot 1.5\,\tfrac{\text{m}}{\text{s}} = 1800\,\text{W}$, like the given power. A speed of $1.5\,\text{m/s}$ is about $5.4\,\text{km/h}$, the pace of brisk walking, which suits a conveyor belt.

8. Exercise 2-8

   Two cranes each lift a $500\,\text{kg}$ load to the same height at a constant velocity. Crane A does the lift in $20\,\text{s}$ and crane B in $40\,\text{s}$.

   Which statement is correct?

   Choose an option.

   • The power of crane B is greater, because it does more work in a longer time.
   • The power of crane A is greater, because the work is the same but it is done in a shorter time.
   • The cranes have the same power, because they lift the same load to the same height.
   • The work of crane A is greater, because it lifts the same load faster.

   Answer
   I have done my solution.
   Answer

   The power of crane A is greater, because the work is the same but it is done in a shorter time.

   Show the solution

   Find

   which crane is more powerful and why

   Principle

   The load and the lifting height are the same, so the lifting work $W = m g h$ is the same for both cranes. Power is work per unit time, $P = \dfrac{W}{t}$, so the same work in a shorter time means a greater power.

   Model answer

   The power of crane A is greater, because the work is the same but it is done in a shorter time. The lifting work is not greater for either, because it depends only on the load and the height, not on the time the lift takes. The power is not the same either, because power tells how fast the work is done: the lift of crane A takes only half the time of crane B, so its power is double. The numbers $500\,\text{kg}$ and the height do not affect the comparison, because the lifting work is the same for both.

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

1. Exercise 2-9Core

   A tractor does $45\,\text{kJ}$ of work by pulling a load with a $1.5\,\text{kN}$ force in the direction of motion. How far does the load move?

   mCheck
   Answer

   $30\,\text{m}$

   Show the solution

   Given

   work $W = 45\,\text{kJ}$, force $F = 1.5\,\text{kN}$ (in the direction of motion)

   Find

   the displacement $s$ of the load in metres

   Principle

   The force is in the direction of motion and constant, so the work is the product of force and distance. From this the distance can be solved when the work and the force are known.

   Formula

   $W = F s$, so $s = \dfrac{W}{F}$

   Conversions

   $45\,\text{kJ} = 45\,000\,\text{J}$ and $1.5\,\text{kN} = 1500\,\text{N}$, so that the units are SI units and $\tfrac{\text{J}}{\text{N}} = \text{m}$.

   Substitution

   $s$$= \dfrac{45\,000\,\text{J}}{1500\,\text{N}}$

   $= 30\,\text{m}$

   Answer

   $30\,\text{m}$

   Two significant figures like the force $1.5\,\text{kN}$.

   Reasonableness

   Units:

   $\dfrac{\text{J}}{\text{N}}$$= \dfrac{\text{N}\cdot\text{m}}{\text{N}}$

   $= \text{m}$

   Reverse check: $1500\,\text{N} \cdot 30\,\text{m} = 45\,000\,\text{J}$, as in the given work. A displacement of thirty metres is believable for a tractor pulling a load.

2. Exercise 2-10Core

   The power of a water pump is $1.5\,\text{kW}$. How much work does the pump do in 4.0 minutes? Give your answer in kilojoules.

   kJCheck
   Answer

   $360\,\text{kJ}$

   Show the solution

   Given

   power $P = 1.5\,\text{kW}$, time $t = 4.0\,\text{min}$

   Find

   the work $W$ done by the pump in kilojoules

   Principle

   Power is work per unit time. The power of the pump stays constant, so the work is power times time.

   Formula

   $P = \dfrac{W}{t}$, so $W = P t$

   Conversions

   $1.5\,\text{kW}$$= 1500\,\text{W}$

   $4.0\,\text{min}$$= 4.0 \cdot 60\,\text{s}$

   $= 240\,\text{s}$

   so that the work comes out in joules. Finally $1\,\text{kJ} = 1000\,\text{J}$.

   Substitution

   $W$$= 1500\,\text{W} \cdot 240\,\text{s}$

   $= 360\,000\,\text{J}$

   $= 360\,\text{kJ}$

   Answer

   $360\,\text{kJ}$

   Two significant figures like the power $1.5\,\text{kW}$ and the time $4.0\,\text{min}$.

   Reasonableness

   Another route: $1.5\,\text{kW} \cdot \tfrac{4}{60}\,\text{h} = 0.10\,\text{kWh}$, and $1\,\text{kWh} = 3600\,\text{kJ}$, so $0.10 \cdot 3600\,\text{kJ} = 360\,\text{kJ}$. This work could lift $1000\,\text{kg}$ of water to a height of about $37\,\text{m}$:

   $\tfrac{360\,000\,\text{J}}{1000\,\text{kg}\cdot 9.81\,\text{m/s}^2}$

   $\approx 37\,\text{m}$
3. Exercise 2-11

   A crane lifts an $850\,\text{kg}$ concrete element slowly to a height of $30\,\text{m}$. How much work does the lifting force do? Give your answer in kilojoules.

   kJCheck
   Answer

   $250\,\text{kJ}$

   Show the solution

   Given

   mass $m = 850\,\text{kg}$, lifting height $h = 30\,\text{m}$, slow lift, $g = 9.81\,\text{m/s}^2$

   Find

   the work $W$ done by the lifting force in kilojoules

   Principle

   In a slow lift the lifting force is equal to the weight and is constant and in the direction of motion. The work is therefore the product of weight and lifting height, and the weight is obtained from the mass.

   Formula

   $W = F h$, $F = G = m g$, so $W = m g h$

   Conversions

   The mass is in kilograms and the height in metres, so no conversions are needed before the calculation. Finally $1\,\text{kJ} = 1000\,\text{J}$.

   Substitution

   $W$$= 850\,\text{kg} \cdot 9.81\,\tfrac{\text{m}}{\text{s}^2} \cdot 30\,\text{m}$

   $= 250\,155\,\text{J}$

   $= 250.155\,\text{kJ}$

   Answer

   $250\,\text{kJ}$

   Two significant figures like the height $30\,\text{m}$.

   Reasonableness

   The weight of the element is

   $850\,\text{kg} \cdot 9.81\,\tfrac{\text{m}}{\text{s}^2}$

   $\approx 8.3\,\text{kN}$

   Rough estimate: $8\,\text{kN} \cdot 30\,\text{m} = 240\,\text{kJ}$, close to the result. A height of thirty metres corresponds to a ten-storey building, and a work of a quarter of a megajoule is believable for a crane.

4. Exercise 2-12

   A cart is pushed by a force in the direction of motion. The force is first $10\,\text{N}$ over the distance $0\text{–}2.0\,\text{m}$. After that the force increases steadily from $10\,\text{N}$ to $30\,\text{N}$ over the distance $2.0\text{–}6.0\,\text{m}$. Calculate the total work done by the force as the area under the $(s, F)$ graph.

   JCheck
   Answer

   $100\,\text{J}$

   Show the solution

   Given

   force $F_1 = 10\,\text{N}$ constant over the distance $s_1 = 2.0\,\text{m}$ ($0\text{–}2.0\,\text{m}$); then the force increases steadily from $F_1 = 10\,\text{N}$ to $F_2 = 30\,\text{N}$ over the distance $s_2 = 4.0\,\text{m}$ ($2.0\text{–}6.0\,\text{m}$)

   Find

   the total work $W$ in joules

   Principle

   The force changes, so the formula $W = F s$ does not work for the whole distance. The work is the area under the $(s,F)$ graph, and it can be calculated in parts: the constant-force part is a rectangle and the steadily increasing part is a trapezium.

   Reading the graph

   The graph has a horizontal segment at the height $10\,\text{N}$ from $0\ldots 2.0\,\text{m}$ and, continuing from it, a straight segment from the height $10\,\text{N}$ to the height $30\,\text{N}$ from $2.0\ldots 6.0\,\text{m}$. The rectangle has height $10\,\text{N}$ and width $2.0\,\text{m}$. The parallel sides of the trapezium are $10\,\text{N}$ and $30\,\text{N}$, and its height (the width along the $s$-axis) is $4.0\,\text{m}$.

   Formula

   $W = A_\text{rectangle} + A_\text{trapezium}$, so $W = F_1 s_1 + 0.5 \cdot (F_1 + F_2)\, s_2$

   Conversions

   The forces are in newtons and the distances in metres, so no conversions are needed.

   Substitution

   $W$$= 10\,\text{N} \cdot 2.0\,\text{m} + \tfrac{1}{2} \cdot (10\,\text{N} + 30\,\text{N}) \cdot 4.0\,\text{m}$

   $= 20\,\text{J} + 80\,\text{J}$

   $= 100\,\text{J}$

   Answer

   $100\,\text{J}$

   Two significant figures like the given values (for example $2.0\,\text{m}$).

   Reasonableness

   The average force over the whole $6.0\,\text{m}$ is

   $\dfrac{100\,\text{J}}{6.0\,\text{m}}$$\approx 17\,\text{N}$

   which lies between the smallest ($10\,\text{N}$) and greatest ($30\,\text{N}$) values of the force. Rough bounds: at least $10\,\text{N} \cdot 6.0\,\text{m} = 60\,\text{J}$ and at most $30\,\text{N} \cdot 6.0\,\text{m} = 180\,\text{J}$, and the result lies between them.

5. Exercise 2-13

   A cart is pulled along a rail by a force in the direction of motion, and the magnitude of the force is measured with a force sensor. The results are in the table (example data set).

   Calculate the work done by the force over the distance $0\text{–}4.0\,\text{m}$ by dividing the distance into four parts of one metre and using, for each part, the average of the force at the start and at the end of the part.

   Force pulling a cart as a function of distance (example data set)

   $s\,(\text{m})$	$F\,(\text{N})$
   0.0	18.0
   1.0	17.2
   2.0	15.1
   3.0	11.0
   4.0	6.0

   JCheck
   Answer

   $55\,\text{J}$

   Show the solution

   Given

   from the table, the force $F$ at $s = 0.0\,\text{m}$, $1.0\,\text{m}$, $2.0\,\text{m}$, $3.0\,\text{m}$ and $4.0\,\text{m}$: $18.0\,\text{N}$, $17.2\,\text{N}$, $15.1\,\text{N}$, $11.0\,\text{N}$ and $6.0\,\text{N}$; the force is in the direction of motion

   Find

   the work $W$ done by the force in joules over the distance $0\ldots 4.0\,\text{m}$

   Principle

   The force changes along the way, so the work is the area under the $(s,F)$ graph. For each one-metre part the area is a trapezium whose height is the average of the force at the ends of the part, and the parts are added together.

   Reading the graph

   The points of the table are thought of as joined by straight segments in the $(s,F)$ coordinate system. At the ends of the part $0\ldots 1.0\,\text{m}$ the force is $18.0\,\text{N}$ and $17.2\,\text{N}$, of the part $1.0\ldots 2.0\,\text{m}$ $17.2\,\text{N}$ and $15.1\,\text{N}$, of the part $2.0\ldots 3.0\,\text{m}$ $15.1\,\text{N}$ and $11.0\,\text{N}$, and of the part $3.0\ldots 4.0\,\text{m}$ $11.0\,\text{N}$ and $6.0\,\text{N}$. The width of each part is $\Delta s = 1.0\,\text{m}$.

   Formula

   $W = \sum 0.5 \cdot (F_\text{start} + F_\text{end})\,\Delta s$, so $W = W_1 + W_2 + W_3 + W_4$

   Conversions

   The forces are in newtons and the distances in metres, so no conversions are needed.

   Substitution

   $$\begin{aligned} W_1 &= 0.5 \cdot (18.0 + 17.2)\,\text{N} \\ &\quad \cdot 1.0\,\text{m} = 17.6\,\text{J} \\ W_2 &= 0.5 \cdot (17.2 + 15.1)\,\text{N} \\ &\quad \cdot 1.0\,\text{m} = 16.15\,\text{J} \\ W_3 &= 0.5 \cdot (15.1 + 11.0)\,\text{N} \\ &\quad \cdot 1.0\,\text{m} = 13.05\,\text{J} \\ W_4 &= 0.5 \cdot (11.0 + 6.0)\,\text{N} \\ &\quad \cdot 1.0\,\text{m} = 8.5\,\text{J} \\ W &= 17.6\,\text{J} + 16.15\,\text{J} \\ &\quad + 13.05\,\text{J} + 8.5\,\text{J} \\ &= 55.3\,\text{J} \end{aligned}$$

   Answer

   $55\,\text{J}$

   Two significant figures, because the width of a part $\Delta s = 1.0\,\text{m}$ and the final force $6.0\,\text{N}$ are given with two significant figures.

   Reasonableness

   The average force over the whole $4.0\,\text{m}$ is $55.3\,\text{J} / 4.0\,\text{m} \approx 14\,\text{N}$, which is between the smallest ($6.0\,\text{N}$) and the greatest ($18.0\,\text{N}$) value of the force. Rough estimate: $18\,\text{N} \cdot 4.0\,\text{m} = 72\,\text{J}$ is an upper limit and $6\,\text{N} \cdot 4.0\,\text{m} = 24\,\text{J}$ a lower limit, and the result is between them.

6. Exercise 2-14

   A student measures the work and power while climbing a staircase. The student's mass is $62\,\text{kg}$. The staircase has 18 steps, each with a height of $0.17\,\text{m}$. The student climbs the stairs three times at as steady a pace as possible, and timing gives the climbing times in the table (example data set).

   Assume that only lifting the student's weight to the height of the staircase requires work. Calculate the student's average power during the climb using the average of the climbing times.

   Stair climbing times (example data set)

   Climb	$t\,(\text{s})$
   1	8.1
   2	7.8
   3	8.4

   WCheck
   Answer

   $230\,\text{W}$

   Show the solution

   Given

   mass $m = 62\,\text{kg}$, 18 steps, step height $0.17\,\text{m}$, climbing times $8.1\,\text{s}$, $7.8\,\text{s}$ and $8.4\,\text{s}$, $g = 9.81\,\text{m/s}^2$

   Find

   the average power $P$ in watts

   Principle

   At a steady pace the lifting force is equal to the student's weight, and the work is calculated only from lifting the weight to the height of the staircase: $W = m g h$. The power is this work divided by the climbing time. The average of the repeated times is used, because the random error of a single timing gets smaller in the average.

   Formula

   $h = n\, h_1$, $W = m g h$ and $P = \dfrac{W}{t_\text{avg}}$, so

   $P$$= \dfrac{m g\, n\, h_1}{t_\text{avg}}$

   Conversions

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

   Substitution

   $$\begin{aligned} \textstyle\sum t &= 8.1\,\text{s} + 7.8\,\text{s} + 8.4\,\text{s} \\ &= 24.3\,\text{s} \\ t_\text{avg} &= \dfrac{24.3\,\text{s}}{3} = 8.1\,\text{s} \\ h &= 18 \cdot 0.17\,\text{m} = 3.06\,\text{m} \\ W &= 62\,\text{kg} \cdot 9.81\,\dfrac{\text{m}}{\text{s}^2} \cdot 3.06\,\text{m} \\ &= 1861.2\,\text{J} \\ P &= \dfrac{1861.2\,\text{J}}{8.1\,\text{s}} \\ &= 229.8\,\text{W} \end{aligned}$$

   Answer

   $230\,\text{W}$

   Two significant figures like the mass $62\,\text{kg}$ and the time $8.1\,\text{s}$.

   Reasonableness

   Rough estimate: the weight is about $600\,\text{N}$ and the height about $3\,\text{m}$, so the work is about $1800\,\text{J}$ and the power $\tfrac{1800\,\text{J}}{8\,\text{s}} \approx 225\,\text{W}$. The range of the three time measurements $7.8\ldots 8.4\,\text{s}$ corresponds to a power of $222\ldots 239\,\text{W}$, so the result $230\,\text{W}$ is within this range. The power of a person climbing stairs quickly is hundreds of watts.

7. Exercise 2-15

   In a problem a $40\,\text{kg}$ box is lifted slowly to a height of $1.5\,\text{m}$ and the work done by the lifting force is asked. A student solved: “$W = F s = 40 \cdot 1.5\,\text{J} = 60\,\text{J}$.”

   The student's result is wrong. Find the error and calculate the correct work in joules.

   JCheck
   Answer

   $590\,\text{J}$

   Show the solution

   Given

   mass $m = 40\,\text{kg}$, lifting height $h = 1.5\,\text{m}$, slow lift, $g = 9.81\,\text{m/s}^2$; the student's result $60\,\text{J}$

   Model answer

   The error: in the formula $W = F s$ the student used the numerical value of the mass, $40$, as the force, although mass is not a force. The lifting force is equal to the weight, $F = m g$, and it must be calculated in newtons. The correct work is $W = m g h \approx 590\,\text{J}$.

   Find

   the student's error and the correct work $W$ done by the lifting force in joules

   Principle

   The student put the number $40$ of the mass in the place of the force in the formula $W = F s$. Mass is not a force: in a slow lift the lifting force is equal to the weight $G = m g$, and it must be brought into the calculation in newtons. The work is the product of this force and the lifting height.

   Formula

   $F = G = m g$ and $W = F h$, so $W = m g h$

   Conversions

   The mass is in kilograms and the height in metres, so no conversions are needed. The unit of weight is the newton: $1\,\text{N} = 1\,\text{kg}\cdot\text{m/s}^2$.

   Substitution

   $$\begin{aligned} G &= 40\,\text{kg} \cdot 9.81\,\dfrac{\text{m}}{\text{s}^2} = 392.4\,\text{N} \\ W &= 392.4\,\text{N} \cdot 1.5\,\text{m} = 588.6\,\text{J} \end{aligned}$$

   Answer

   $590\,\text{J}$

   Two significant figures like the mass $40\,\text{kg}$ and the height $1.5\,\text{m}$.

   Reasonableness

   The student's result is almost a tenth of the correct one, which corresponds to the factor $g \approx 9.81\,\text{m/s}^2$ that was left out. Rough estimate: the weight is about $400\,\text{N}$, so $400\,\text{N} \cdot 1.5\,\text{m} = 600\,\text{J}$. Units: the student's number $40$ was a mass in kilograms and not in newtons, so the product could not be in joules.

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

1. Exercise 2-16

   A crate is pushed by a force in the direction of motion. The force increases steadily from zero to $40\,\text{N}$ over the distance $0\text{–}0.50\,\text{m}$, then stays at $40\,\text{N}$ over the distance $0.50\text{–}1.5\,\text{m}$ and decreases steadily to zero over the distance $1.5\text{–}2.0\,\text{m}$.

   What is the average magnitude of the force with respect to distance over the whole $2.0\,\text{m}$? Hint: first calculate the total work as the area under the graph.

   NCheck
   Answer

   $30\,\text{N}$

   Show the solution

   Given

   peak force $F = 40\,\text{N}$; the force increases from zero to $F$ over the distance $s_1 = 0.50\,\text{m}$, stays constant over the distance $s_2 = 1.0\,\text{m}$ and decreases to zero over the distance $s_3 = 0.50\,\text{m}$; total distance $s = 2.0\,\text{m}$

   Find

   the average magnitude of the force with respect to distance, $F_\text{avg}$, in newtons

   Principle

   The force changes, so the work is the area under the $(s,F)$ graph. The average force is the constant force that would do the same work over the same distance, so $W = F_\text{avg}\, s$.

   Reading the graph

   The graph is a trapezium: a rising triangle ($0\ldots 0.50\,\text{m}$, height $40\,\text{N}$), a horizontal part as a rectangle ($0.50\ldots 1.5\,\text{m}$, height $40\,\text{N}$) and a falling triangle ($1.5\ldots 2.0\,\text{m}$, height $40\,\text{N}$). The work is the sum of these three areas.

   Formula

   $W = \tfrac{1}{2} F s_1 + F s_2 + \tfrac{1}{2} F s_3$ and $W = F_\text{avg}\, s$, so $F_\text{avg} = \dfrac{W}{s}$

   Conversions

   The forces are in newtons and the distances in metres, so no conversions are needed.

   Substitution

   $W$$= \tfrac{1}{2} \cdot 40\,\text{N} \cdot 0.50\,\text{m} + 40\,\text{N} \cdot 1.0\,\text{m} + \tfrac{1}{2} \cdot 40\,\text{N} \cdot 0.50\,\text{m}$

   $= 10\,\text{J} + 40\,\text{J} + 10\,\text{J}$

   $= 60\,\text{J}$

   $F_\text{avg}$$= \dfrac{60\,\text{J}}{2.0\,\text{m}}$

   $= 30\,\text{N}$

   Answer

   $30\,\text{N}$

   Two significant figures like the distance $2.0\,\text{m}$.

   Reasonableness

   The average is smaller than the peak value $40\,\text{N}$, because the force is small at the beginning and at the end, but greater than half of the peak, because the force is at its peak for half of the distance. The trapezium formula gives the same work: $0.5 \cdot (2.0\,\text{m} + 1.0\,\text{m}) \cdot 40\,\text{N} = 60\,\text{J}$. The average of a force increasing steadily from zero would be $20\,\text{N}$.

2. Exercise 2-17

   In Joule's experiment a $2.0\,\text{kg}$ weight falls $1.5\,\text{m}$ and turns a paddle wheel in water. The experiment is repeated 20 times, each time lifting the weight back up. Assume that as it falls the weight loses an amount of energy equal to the work of lifting it, and that all of this energy stays in the water.

   By how much does the internal energy of the water increase?

   JCheck
   Answer

   $590\,\text{J}$

   Show the solution

   Given

   mass $m = 2.0\,\text{kg}$, falling height $h = 1.5\,\text{m}$, number of drops $n = 20$, $g = 9.81\,\text{m/s}^2$

   Find

   the increase in the internal energy of the water $\Delta U$ in joules

   Principle

   Energy is conserved. As it falls, the weight loses an amount of energy equal to the lifting work $m g h$, and by assumption all of this energy is transferred through the paddle wheel to the water as internal energy. Each drop transfers the same energy, so the drops are added together.

   Formula

   $W_1 = m g h$ and $\Delta U = n W_1$, so $\Delta U = n m g h$

   Conversions

   The mass is in kilograms and the height in metres, so no conversions are needed. The number of drops $n = 20$ is just a number.

   Substitution

   $W_1$$= 2.0\,\text{kg} \cdot 9.81\,\tfrac{\text{m}}{\text{s}^2} \cdot 1.5\,\text{m}$

   $= 29.43\,\text{J}$

   $\Delta U$$= 20 \cdot 29.43\,\text{J}$

   $= 588.6\,\text{J}$

   Answer

   $590\,\text{J}$

   Two significant figures like the mass $2.0\,\text{kg}$ and the height $1.5\,\text{m}$.

   Reasonableness

   Rough estimate: the weight is about $20\,\text{N}$, so one drop gives about $20\,\text{N} \cdot 1.5\,\text{m} = 30\,\text{J}$ and 20 drops about $600\,\text{J}$. The result is the same as lifting an object of about $60\,\text{kg}$ by a metre, a small amount of energy, so the temperature of the water changed only a little and the measurement needed accuracy. That is why Joule repeated the drops many times.

3. Exercise 2-18

   A solar panel at a cabin gives a power of $180\,\text{W}$ in sunshine.

   For how many minutes must the sun shine on the panel so that the energy is enough to lift $1000\,\text{kg}$ of water to a height of $50\,\text{m}$? Assume that all the energy goes into lifting the water.

   minCheck
   Answer

   $45\,\text{min}$

   Show the solution

   Given

   power of the panel $P = 180\,\text{W}$, mass of water $m = 1000\,\text{kg}$, lifting height $h = 50\,\text{m}$, $g = 9.81\,\text{m/s}^2$, all the energy goes into lifting

   Find

   the time $t$ in minutes

   Principle

   By assumption the energy produced by the panel is equal to the work needed to lift the water, and there are no losses. The power is constant, so the energy produced is power times time. The work of lifting is $m g h$, because the water is lifted slowly.

   Formula

   $W = m g h$ and $W = P t$, so $t = \dfrac{m g h}{P}$

   Conversions

   The mass, height and power are in SI units, so the time comes out first in seconds. Finally $1\,\text{min} = 60\,\text{s}$.

   Substitution

   $W$$= 1000\,\text{kg} \cdot 9.81\,\tfrac{\text{m}}{\text{s}^2} \cdot 50\,\text{m}$

   $= 490\,500\,\text{J}$

   $t$$= \dfrac{490\,500\,\text{J}}{180\,\text{W}}$

   $= 2725\,\text{s}$

   $= \dfrac{2725}{60}\,\text{min}$

   $= 45.42\,\text{min}$

   Answer

   $45\,\text{min}$

   Two significant figures like the height $50\,\text{m}$.

   Reasonableness

   In one hour the panel produces $180\,\text{W} \cdot 3600\,\text{s} = 648\,\text{kJ}$, which is more than the required $490\,\text{kJ}$, so the time is less than an hour. The ratio $\tfrac{490.5}{648} \approx 0.76$ hours is about $45\,\text{min}$, as in the result. This matches everyday experience of solar electricity.

4. Exercise 2-19

   In a spring stretching experiment the spring is stretched slowly, and a force sensor measures the stretching force $F$ along the spring at different extensions $s$ (example data set). The extension is measured from the free length of the spring.

   a) Use the values in the table to justify how the force depends on the extension, and explain why the work done in stretching is the area of a triangle in the $(s,F)$ graph.

   b) Calculate the work when the spring is stretched to an extension of $10.0\,\text{cm}$.

   c) To what height could this work slowly lift an object of $0.20\,\text{kg}$? Justify the model of the calculation and give the result of part c) in centimetres in the answer box.

   Stretching force of a spring as a function of extension (example data set)

   $s\,(\text{cm})$	$F\,(\text{N})$
   2.0	1.9
   4.0	4.2
   6.0	5.8
   8.0	8.1
   10.0	10.0

   cmCheck
   Answer

   $25\,\text{cm}$

   Show the solution

   Given

   from the table $(s, F)$: $(2.0\,\text{cm}$; $1.9\,\text{N})$, $(4.0\,\text{cm}$; $4.2\,\text{N})$, $(6.0\,\text{cm}$; $5.8\,\text{N})$, $(8.0\,\text{cm}$; $8.1\,\text{N})$, $(10.0\,\text{cm}$; $10.0\,\text{N})$; extension $s_\text{max} = 10.0\,\text{cm}$; mass to be lifted $m = 0.20\,\text{kg}$, $g = 9.81\,\text{m/s}^2$

   Find

   a) how the force depends on the extension, b) the work $W$ done in stretching, c) the lifting height $h$ in centimetres

   Principle

   The force changes with the extension, so the work is the area under the $(s,F)$ graph. If the force is proportional to the extension, the graph is a straight line through the origin and the area is a triangle. Energy is conserved without losses: the work done in stretching the spring can be used to lift an object slowly, so the lifting work $m g h$ is equal to $W$.

   Reading the graph

   The force grows with the extension so that the quotient $\tfrac{F}{s}$ is about $1.0\,\text{N/cm}$ at all points: for example $\tfrac{1.9}{2.0} = 0.95$ and $\tfrac{10.0}{10.0} = 1.0$. Because of measurement uncertainty the measured points are usually not exactly on one line, so a line through the origin is fitted among them; the origin belongs on the line, because the extension is measured from the free length. On the line, at $s = 10.0\,\text{cm}$ the force is $F = 10.0\,\text{N}$. Because the force grows steadily from zero, the graph is a straight line through the origin and the area under it is a triangle; therefore the work is the area of a triangle whose base is $s$ and whose height is $F$.

   Formula

   $W = \tfrac{1}{2} F s$ and $W = m g h$, so $h = \dfrac{W}{m g}$

   Conversions

   $10.0\,\text{cm} = 0.100\,\text{m}$, so that the work comes out in joules. Finally $h$ is converted to centimetres: $1\,\text{m} = 100\,\text{cm}$.

   Substitution

   $$\begin{aligned} W &= 0.5 \cdot 10.0\,\text{N} \cdot 0.100\,\text{m} \\ &= 0.500\,\text{J} \\ h &= \dfrac{0.500\,\text{J}}{0.20\,\text{kg} \cdot 9.81\,\dfrac{\text{m}}{\text{s}^2}} = 0.2548\,\text{m} \\ &= 25.48\,\text{cm} \end{aligned}$$

   Answer

   a) the force is approximately proportional to the extension, $F \approx 1.0\,\tfrac{\text{N}}{\text{cm}} \cdot s$; b) $W = 0.50\,\text{J}$; c) $25\,\text{cm}$ ($25.5\,\text{cm}$ before rounding), two significant figures like the mass $0.20\,\text{kg}$.

   Reasonableness

   If the force were $10.0\,\text{N}$ over the whole distance, the work would be $10.0\,\text{N} \cdot 0.100\,\text{m} = 1.0\,\text{J}$; the work of the triangle is half of this, because the average force is $5.0\,\text{N}$. The trapezium sum of the measured points gives the same: $0.020\,\text{m} \cdot 25.0\,\text{N} = 0.50\,\text{J}$. The weight of the object is

   $0.20\,\text{kg} \cdot 9.81\,\tfrac{\text{m}}{\text{s}^2}$

   $\approx 2.0\,\text{N}$

   and $2.0\,\text{N} \cdot 0.25\,\text{m} \approx 0.50\,\text{J}$.

5. Exercise 2-20

   The motor of a construction crane lifts a $450\,\text{kg}$ load from the floor upwards. The height of the load above the floor is measured at different moments, counting from the start of the lifting motion (example data set).

   a) Justify from the table that the load rises at approximately constant velocity, and determine the velocity using a fitted line. The line may be drawn through the origin, because time and height are counted from the start of the lifting motion.

   b) Justify why the lifting force is equal to the weight of the load when the load rises at constant velocity (as in the derivation of the formula $W = m g h$).

   c) Calculate the power the motor gives to the load in kilowatts. This result goes in the answer box.

   d) Calculate the lifting work $m g h$ for the last point of the table ($8.0\,\text{s}$; $6.5\,\text{m}$) and compare it with the work obtained as the product of the power and the time $8.0\,\text{s}$.

   Height of the lifted load (example data set)

   $t\,(\text{s})$	$h\,(\text{m})$
   0.0	0.0
   2.0	1.6
   4.0	3.3
   6.0	4.9
   8.0	6.5

   kWCheck
   Answer

   $3.6\,\text{kW}$

   Show the solution

   Given

   mass $m = 450\,\text{kg}$, from the table $(t, h)$: $(2.0\,\text{s}$; $1.6\,\text{m})$, $(4.0\,\text{s}$; $3.3\,\text{m})$, $(6.0\,\text{s}$; $4.9\,\text{m})$, $(8.0\,\text{s}$; $6.5\,\text{m})$, $g = 9.81\,\text{m/s}^2$

   Find

   a) the velocity $v$ of the load, b) why the lifting force is equal to the weight, c) the power $P$ of the motor in kilowatts, d) a comparison of the work at the last point of the table

   Principle

   If the height grows steadily with time, the load rises at constant velocity. At constant velocity the lifting force $F$ is equal to the weight $G = m g$, and the power is the product of force and velocity, $P = F v$.

   Reading the graph

   In time intervals of $2.0\,\text{s}$ the height grows by $1.6\,\text{m}$, $1.7\,\text{m}$, $1.6\,\text{m}$ and $1.6\,\text{m}$, that is, by almost the same amount, so the velocity is approximately constant. Because of measurement uncertainty the measured points are usually not exactly on one line, so the velocity is read from the slope of a fitted line, not from two adjacent measured points. A line through the origin fits among the points, because time and height are counted from the start of the lifting motion. From the line $h = 6.52\,\text{m}$ is read at $t = 8.0\,\text{s}$, so the slope is

   $v$$= 6.52\,\text{m} / 8.0\,\text{s}$

   $= 0.815\,\text{m/s}$

   Formula

   $F = G = m g$ and $P = F v$, so $P = m g v$; in the comparison $W = P t$ and $W = m g h$

   Conversions

   All quantities are in SI units. Finally $1\,\text{kW} = 1000\,\text{W}$.

   Substitution

   $$\begin{aligned} G &= 450\,\text{kg} \cdot 9.81\,\dfrac{\text{m}}{\text{s}^2} \\ &= 4414.5\,\text{N} \\ P &= 4414.5\,\text{N} \cdot 0.815\,\dfrac{\text{m}}{\text{s}} = 3598\,\text{W} \\ &= 3.598\,\text{kW} \\ P t &= 3598\,\text{W} \cdot 8.0\,\text{s} \\ &= 28\,780\,\text{J} \\ m g h &= 4414.5\,\text{N} \cdot 6.5\,\text{m} \\ &= 28\,690\,\text{J} \end{aligned}$$

   Answer

   $3.6\,\text{kW}$

   Two significant figures like the mass and the height measurements. Justifications: a) the height grows steadily, so the velocity is approximately constant, $v \approx 0.82\,\text{m/s}$; b) as the lesson says, at constant velocity the lifting force is equal to the weight, as in the derivation of the formula $W = m g h$; d) the works are $28.8\,\text{kJ}$ and $28.7\,\text{kJ}$, that is, they differ by less than one percent.

   Reasonableness

   Rough estimate: the weight is about $4400\,\text{N}$ and the velocity about $0.8\,\text{m/s}$, so the power is about $3500\,\text{W}$. The comparison d) is consistent: the work $P t$ given by the fitted line and the work $m g h$ calculated from the point of the table differ by less than one percent. A power of a few kilowatts is believable for the lifting motor of a construction crane when only the lifting work is taken into account.