CAIE A-Level Physics AS 5 Work Energy and Power Questions

Practise analysing work, energy transfers, efficiency and power, applying gravitational and kinetic energy relationships and using P = W/t or P = Fv.

Exam points

calculate work and track energy transfers using work, conservation of energy, efficiency and power
derive and apply gravitational potential and kinetic energy relationships in changing-motion problems
use P = W/t and P = Fv to analyse rates of energy transfer and mechanical performance

Covered syllabus points

5.1 Energy conservation 5.2 Gravitational potential energy and kinetic energy

[Maximum number: 5]

(a)

A child's toy uses the spring in (a) to launch a ball of mass 0.020 kg vertically into the air. The ball is initially held against one end of the spring which has a compression of 0.045 m . The spring is then released to launch the ball. The kinetic energy of the ball as it leaves the toy is 0.72 J .

[ 1 ]

(i)

The toy converts the elastic potential energy of the spring into the kinetic energy of the ball. Use the information in (a)(ii) to calculate the percentage efficiency of this conversion.

[ 1 ]

$\begin{aligned} \text { efficiency } =(0.72 / 0.81) \times 100 =89 \% \end{aligned}$

(b)

The ball in (b) leaves the toy at point A and moves vertically upwards through the air. Point B is the position of the ball when it is at maximum height h above point A , as illustrated in Fig. 3.2.

Fig. 3.2 (not to scale)

The gravitational potential energy of the ball increases by 0.60 J as it moves from A to B .

[ 4 ]

(i)

Calculate h.

[ 2 ]

$(\Delta) E=m g(\Delta) h$ $h=0.60 /(0.020 \times 9.81)=3.1 \mathrm{~m}$

(ii)

Determine the average force due to air resistance acting on the ball for its movement from A to B.
average force = N

[ 2 ]

F=(0.72-0.60) / 3.1
$=0.039 \mathrm{~N}$

[Maximum number: 9]

(a)

A mass m moves a vertical distance $\Delta h$ in a uniform gravitational field and gains gravitational potential energy $\Delta E_{\mathrm{p}}$ . The acceleration of free fall is g.

Use the concept of work done to show that

\Delta E_{\mathrm{P}}=m g \Delta h .

[ 2 ]

work (done) = force × displacement
( force =m g and distance $=\Delta h)$

(so) work (done) $=m g \Delta h$ and work $=\Delta E_{(\mathrm{P})}$ (so $\Delta \mathrm{E}_{(\mathrm{P})}=m g \Delta h$ )

(b)

A 0.60 kg mass is attached to a string which is wrapped around the wheel of a generator, as shown in Fig. 4.1.

Fig. 4.1

The mass is held stationary above the floor. When released, the mass initially accelerates and then falls at a steady speed and spins the wheel. The generator causes a current in a resistor. Air resistance is negligible.

State the main energy change when the mass is falling at a steady speed. energy to energy.

[ 1 ]

(c)

When falling at a steady speed, the mass in (b) falls through a vertical distance of 1.4 m in a time of 4.0 s . This causes a current of 90 mA in the resistor. The resistance of the resistor is $47 \Omega$ .

Calculate:

[ 6 ]

(i)

the rate of work done by the falling mass
rate of work done = W

[ 2 ]

$P=m g(\Delta) h /(\Delta) t$ or F v
$\begin{aligned} P =(0.60 \times 9.81 \times 1.4) / 4.0 \text { or } 0.60 \times 9.81 \times(1.4 / 4.0) =2.1 \mathrm{~W} \end{aligned}$

(ii)

the power dissipated in the resistor
power = W

[ 2 ]

$P=I^{2} R$ or I V or $V^{2} / R$ $\begin{aligned} =0.09^{2} \times 47 \text { or } 0.09 \times 4.23 \text { or } 4.23^{2} / 47 =0.38 \mathrm{~W} \end{aligned}$

(iii)

the efficiency of the generator.
efficiency =

[ 2 ]

efficiency $=P_{\text {out }} / P_{\text {in }}(\times 100)$ or $E_{\text {out }} / E_{\text {in }}(\times 100)$ $\begin{aligned} =0.38 / 2.1(\times 100) \text { or } 0.38 \times 4.0 / 2.1 \times 4.0(\times 100) =0.18 \text { or } 18 \% \end{aligned}$

[Maximum number: 1]

Another parcel is accidentally released from rest by a different aircraft when it is hovering at a great height above the ground. Air resistance is now significant.

Describe the energy conversion that occurs when the parcel is falling through the air at constant (terminal) speed.

[Maximum number: 5]

A pendulum consists of a solid sphere suspended by a string from a fixed point P , as shown in Fig. 3.1.

Fig. 3.1 (not to scale)

The sphere swings from side to side. At one instant the sphere is at its lowest position X , where it has kinetic energy 0.86 J and momentum 0.72 Ns in a horizontal direction. A short time later the sphere is at position Y , where it is momentarily stationary at a maximum vertical height h above position X.

The string has a fixed length and negligible weight. Air resistance is also negligible.

(a)

Show that the mass of the sphere is 0.30 kg .

[ 3 ]

p=m v or 0.72=m v
$E=\frac{1}{2} m v^{2}$ or $0.86=1 / 2 m v^{2}$ $(m=) 0.72^{2} /(2 \times 0.86)=0.30(\mathrm{~kg})$

or
$\begin{aligned} & v=2 E_{\mathrm{K}} / p & v=(0.86 \times 2) / 0.72=2.4 \text { (to } 2 \text { s.f.) } & m=0.72 / 2.4=0.30(\mathrm{~kg}) \end{aligned}$

(b)

Calculate height h.

\begin{aligned} & h= \\ & \text { m [2] } \end{aligned}

[ 2 ]

$(\Delta) E=m g(\Delta) h$ $\begin{aligned} h =0.86 /(0.30 \times 9.81) =0.29 \mathrm{~m} \end{aligned}$

All question bank results loaded

Covered syllabus points

Question 1

Question (a)

Question (i)

Question (b)

Question (i)

Question (ii)

Question 2

Question (a)

Question (b)

Question (c)

Question (i)

Question (ii)

Question (iii)

Question 3

Question 4

Question (a)

Question (b)