Chapter 2: Review

The chapter problem, solved

The bike-hire workshop checks the same numbers on every cargo bike. Take $g = 9.8\,\text{N/kg}$ .

1. The weight of the bike and rider. Their mass is 110 kg.

\begin{aligned} W &= m g \\ &= 110 \times 9.8 \\ &= 1078\,\text{N} \\ &= 1.1 \times 10^3\,\text{N} \quad (2\ \text{significant figures}) \end{aligned}

2. The spring constant of the front spring. The spring compresses by 0.020 m under a force of 600 N.

\begin{aligned} k &= \frac{F}{e} \\ &= \frac{600}{0.020} \\ &= 30000\,\text{N/m} \\ &= 3.0 \times 10^4\,\text{N/m} \end{aligned}

3. The energy stored in the front spring.

\begin{aligned} E_e &= \tfrac{1}{2} k e^2 \\ &= 0.5 \times 30000 \times 0.020^2 \\ &= 6.0\,\text{J} \end{aligned}

This is also the work done in compressing the spring.

4. The work done pushing the bike up the ramp. The force is 150 N along a ramp 8.0 m long.

\begin{aligned} W &= F s \\ &= 150 \times 8.0 \\ &= 1200\,\text{J} \\ &= 1.2 \times 10^3\,\text{J} \end{aligned}

5. The pressure in the front tyre. The contact area is $12\,\text{cm}^2 = 0.0012\,\text{m}^2$ .

\begin{aligned} p &= \frac{F}{A} \\ &= \frac{540}{0.0012} \\ &= 450000\,\text{Pa} \\ &= 4.5 \times 10^5\,\text{Pa} \end{aligned}

6. The moment on the brake lever. The hand pulls with 60 N at 0.090 m from the pivot.

\begin{aligned} M &= F d \\ &= 60 \times 0.090 \\ &= 5.4\,\text{N m} \end{aligned}

7. The force in the brake cable. The cable is 0.030 m from the pivot, so it must give the same moment.

\begin{aligned} F &= \frac{M}{d} \\ &= \frac{5.4}{0.030} \\ &= 180\,\text{N} \end{aligned}

8. Does the lever make the brakes work better? Yes. The rider pulls the lever with 60 N, and the cable pulls with 180 N, which is three times as large. The lever is a force multiplier: the effort acts further from the pivot than the cable, so the same moment gives a larger force at the shorter distance. A larger force in the cable presses the brake blocks harder onto the wheel.

This final judgement is a reasoned answer. It uses the numbers above, and it matches the rule for balanced moments. It is not the kind of claim a computer can check, so compare your reasoning with the evidence in the working.

Mixed practice — choose the method first

For each question, say which equation or idea you will use before you calculate. Keep your units with your numbers.

In the chapter problem the bike and its rider have a mass of 110 kg. Take $g = 9.8\,\text{N/kg}$ . Calculate their total weight.
The front suspension spring of the bike compresses by 0.020 m when the front wheel carries a force of 600 N. Calculate the spring constant of the spring.
The spring constant is $3.0 \times 10^4\,\text{N/m}$ and the spring is compressed by 0.020 m. Calculate the energy stored in the elastic potential store.
A force of 150 N pushes the bike 8.0 m up the workshop ramp, along the line of the force. Calculate the work done.
The front tyre carries a force of 540 N. The contact patch has an area of 12 cm $^2$ . Calculate the pressure in the tyre in pascals.
A hand pulls the brake lever with a force of 60 N at a perpendicular distance of 0.090 m from the pivot. Calculate the moment of the force about the pivot.
The brake cable is 0.030 m from the pivot. The moment about the pivot is 5.4 N m. Calculate the force in the cable.
The bike and its rider have a total mass of 110 kg and travel at 6.0 m/s. Calculate the energy in the kinetic store.
A workshop motor does 1200 J of work, and 900 J of that work is useful. Calculate the efficiency of the motor as a decimal and as a percentage.
A cyclist pedals with a forward force of 260 N. Air resistance and friction give a total backward force of 190 N. Calculate the resultant force on the bicycle.
The bike is lifted 0.50 m onto a rack. Its mass is 110 kg. Take $g = 9.8\,\text{N/kg}$ . Calculate the work done against gravity, using the gravitational potential energy gained.
The chapter problem asks for the pressure in the front tyre and the moment on the brake lever. Say which equation you would use for each part, and name the quantities you need.
H A water tank is 1.2 m deep. The density of water is 1000 kg/m $^3$ and $g = 9.8\,\text{N/kg}$ . Calculate the pressure due to the water at the bottom of the tank.

Answers and concise solutions

1. Total weight of bike and rider

Answer: $W = mg = 110 \times 9.8 = 1078\,\text{N}$ , so $1.1 \times 10^3\,\text{N}$ to 2 significant figures.

2. Spring constant of the front spring

Answer: $k = \dfrac{F}{e} = \dfrac{600}{0.020} = 30000\,\text{N/m} = 3.0 \times 10^4\,\text{N/m}$ .

3. Energy stored in the spring

Answer: $E_e = \tfrac{1}{2}ke^2 = 0.5 \times 30000 \times 0.020^2 = 6.0\,\text{J}$ .

4. Work done pushing the bike up the ramp

Answer: $W = Fs = 150 \times 8.0 = 1200\,\text{J} = 1.2 \times 10^3\,\text{J}$ .

5. Pressure in the front tyre

Answer: First convert the area: $12\,\text{cm}^2 = 0.0012\,\text{m}^2$ . Then $p = \dfrac{F}{A} = \dfrac{540}{0.0012} = 450000\,\text{Pa} = 4.5 \times 10^5\,\text{Pa}$ .

6. Moment of the hand force

Answer: $M = Fd = 60 \times 0.090 = 5.4\,\text{N m}$ .

7. Force in the brake cable

Answer: $F = \dfrac{M}{d} = \dfrac{5.4}{0.030} = 180\,\text{N}$ . The lever multiplies the 60 N hand force three times.

8. Energy in the kinetic store

Answer: $E_k = \tfrac{1}{2}mv^2 = 0.5 \times 110 \times 6.0^2 = 1980\,\text{J}$ , so $2.0 \times 10^3\,\text{J}$ to 2 significant figures.

9. Efficiency of the motor

Answer: $\text{efficiency} = \dfrac{900}{1200} = 0.75 = 75\%$ .

10. Resultant force on the bicycle

Answer: The forces act in opposite directions, so subtract: $260 - 190 = 70\,\text{N}$ forwards.

11. Work done lifting the bike

Answer: $E_p = mgh = 110 \times 9.8 \times 0.50 = 539\,\text{J}$ , so $540\,\text{J}$ to 2 significant figures. The work done against gravity is the same size.

12. Choosing the equation for each part

For the tyre, use $p = \dfrac{F}{A}$ . I need the force on the tyre and the area of the contact patch.
For the brake lever, use $M = Fd$ . I need the force on the lever and the perpendicular distance from the pivot.

13. Pressure at the bottom of the tank

Answer: $p = h\rho g = 1.2 \times 1000 \times 9.8 = 11760\,\text{Pa} = 1.2 \times 10^4\,\text{Pa}$ .

What you can now do

describe forces as interactions, and sort them into contact and non-contact forces;
calculate weight from mass, and find the resultant of two forces in a straight line;
resolve a force into two components, and use a free body diagram or a vector diagram (Higher tier);
use the force-extension relationship and calculate the energy stored in a spring;
calculate work done, force or distance, and describe the energy transfer;
calculate pressure, force or area, and explain atmospheric pressure;
calculate the pressure at a depth in a liquid and explain upthrust (Higher tier);
calculate the moment of a force, use balanced moments, and explain how levers and gears work.

Your relevance reflection

Return to the prompt from the chapter opener: where in your own day do you push, pull or turn something? Write one more example now that you have finished the chapter. Has the chapter changed what you notice about a tool, a bike or a door?