**Exam Style Questions for IBDP Physics 2025 – A.3 Work, energy and power SL Paper 1**

## IBDP Physics SL Paper 1 – All Chapters

## Topic: A.3 Work, Energy and Power SL Paper 1

Work, Kinetic Energy, Gravitational and Elastic Potential Energy, Power, Conservation of Energy, Efficiency

#### Question- A.3 Work, energy and power SL Paper 1

The input power of an electric motor is \(200 \mathrm{~W}\). It is used to raise a mass of \(10 \mathrm{~kg}\) at constant speed. If the efficiency of the motor is \(40 \%\), through what height will the mass be raised in 1 second?

A. \(0.5 \mathrm{~m}\)

B. \(0.8 \mathrm{~m}\)

C. \(1.2 \mathrm{~m}\)

D. \(2.0 \mathrm{~m}\)

**▶️Answer/Explanation**

Ans:B

\(\text{Useful power} = \text{Efficiency} \times \text{Input power}\)

\(\text{Useful power} = 0.40 \times 200 \, \text{W} = 80 \, \text{W}\)

\(\text{Work (W)} = \text{Power (P)} \times \text{Time (t)}\)

In this case, Power (P) is 80 W, and Time (t) is 1 second.

\(\text{Work (W)} = 80 \, \text{W} \times 1 \, \text{s} = 80 \, \text{J} \, (\text{joules})\)

The work done is equal to the change in potential energy of the mass. Use the formula for gravitational potential energy:

\(\text{Gravitational Potential Energy (U)} = mgh\)

Rearrange the formula to solve for \(h):

\(h = \frac{W}{mg}\)

\(h = \frac{80 \, \text{J}}{10 \, \text{kg} \times 9.81 \, \text{m/s}^2} \approx 0.815 \, \text{m}\)

Rounded to one decimal place, the height to which the mass is raised in 1 second is approximately \(0.8 \, \text{m}\).

### Question

A ball falls with terminal velocity through air. What is correct about the kinetic energy and the total energy of the ball?

**▶️Answer/Explanation**

Ans:A

When a ball falls with terminal velocity through air:

- Kinetic energy remains constant because the ball is moving at a constant velocity.
- Total energy decreases.

The decrease in total energy is due to the conversion of potential energy (when the ball is higher) into other forms of energy (such as work done against air resistance and heat dissipation).

So, the correct statement is:

- Kinetic energy: constant
- Total energy: decreases

### Question

Wind generator \(X\) has a maximum power output \(P_X\) for a particular wind speed. For the same wind speed, wind generator \(\mathrm{Y}\) has a maximum power output \(P_{\mathrm{Y}}\).

The blade radius of \(Y\) is three times the blade radius of \(X\). \(Y\) is twice as efficient as \(X\).

What is \(\frac{P_{\mathrm{Y}}}{P_{\mathrm{X}}}\) ?

A. \(\frac{3}{2}\)

B. \(\frac{9}{2}\)

C. $6$

D. $18$

**▶️Answer/Explanation**

Ans:D

Let \(P_X\) be the maximum power output of wind generator \(X\) and \(P_Y\) be the maximum power output of wind generator \(Y\).

We are given that:

1. The blade radius of \(Y\) (\(R_Y\)) is three times the blade radius of \(X\) (\(R_X\)), so \(R_Y = 3R_X\).

2. \(Y\) is twice as efficient as \(X\), so the efficiency of \(Y\) (\(\eta_Y\)) is \(2\) times the efficiency of \(X\) (\(\eta_X\)), or \(\eta_Y = 2\eta_X\).

The power output of a wind generator is given by:

\[P = \frac{1}{2} \rho A v^3 \eta\]

Where:

– \(P\) is the power output.

– \(\rho\) is the air density.

– \(A\) is the swept area of the blades.

– \(v\) is the wind speed.

– \(\eta\) is the efficiency.

The swept area (\(A\)) is proportional to the square of the blade radius (\(R\)), so \(A \propto R^2\).

Now, let’s compare the power outputs of \(X\) and \(Y\) for the same wind speed (\(v\)):

For \(X\):

\[P_X \propto \frac{1}{2} \rho (A_X) v^3 \eta_X\]

\[P_X \propto \frac{1}{2} \rho (R_X^2) v^3 \eta_X\]

For \(Y\):

\[P_Y \propto \frac{1}{2} \rho (A_Y) v^3 \eta_Y\]

\[P_Y \propto \frac{1}{2} \rho (R_Y^2) v^3 \eta_Y\]

Substituting the values of \(R_Y\) and \(\eta_Y\) in terms of \(R_X\) and \(\eta_X\):

\[P_Y \propto \frac{1}{2} \rho ((3R_X)^2) v^3 (2\eta_X)\]

\[P_Y \propto \frac{1}{2} \rho (9R_X^2) v^3 (2\eta_X)\]

Now, let’s compare \(\frac{P_Y}{P_X}\):

\[\frac{P_Y}{P_X} = \frac{\frac{1}{2} \rho (9R_X^2) v^3 (2\eta_X)}{\frac{1}{2} \rho (R_X^2) v^3 \eta_X}\]

The terms \(\frac{1}{2}\rho v^3\) and \(\eta_X\) cancel out:

\[\frac{P_Y}{P_X} = \frac{9R_X^2 \cdot 2\eta_X}{R_X^2 \cdot \eta_X}\]

\[\frac{P_Y}{P_X} = \frac{18\eta_X}{\eta_X} = 18\]

So, \(\frac{P_Y}{P_X} = 18\).

An increasing force acts on a metal wire and the wire extends from an initial length *l*_{0} to a new length *l*. The graph shows the variation of force with length for the wire. The energy required to extend the wire from *l*_{0} to *l *is *E*.

The wire then contracts to half its original extension.

What is the work done by the wire as it contracts?

A. 0.25*E*

B. 0.50*E*

C. 0.75*E*

*D. E *

**Answer/Explanation**

**Markscheme**

**C**

From graph force is linear function of length.

\(F\propto x\) ( \(x\rightarrow\) extension of wire)

\(F=kx\)

let total extension is \(X_o\)

\(work done \Rightarrow w=\int_{0}^{x_o}F.dx\)

\(w=\frac{kx_o^2}{2}=E\) (given)

Now for the half extension.

\(w=\frac{k}{2}\int_{\frac{x_o}{2}}^{x_o}(x_o^2)\)

\(=\frac{k}{2}(x_o^2-\frac{x_o^2}{4})\)

\(=\frac{3}{4}(\frac{kx_o^2}{2})\)

\(=\frac{3}{4}E\)

A motor of input power 160 W raises a mass of 8.0 kg vertically at a constant speed of 0.50 m s^{–1}.

What is the efficiency of the system?

A. 0.63%

B. 25%

C. 50%

D. 100%

**Answer/Explanation**

**Markscheme**

**B**

\(Input power=160W\)

\(Output power=F.V\)

\(=(mg).V\)

\(=(8\times 10).\frac{5}{10}\)

\(P_o=40W\)

\(\eta =\frac{P_o}{P_i}\times 100%\)

\(=\frac{40}{160}\times 100%\)

\(Efficiency=25%\)