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

## IBDP Physics 2025 HL Paper 1 – All Chapters

## A.3 Work, energy and power HL Paper 1

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

### Question-A.3 Work, energy and power

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

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

### Question

A spring of negligible mass is compressed and placed between two stationary masses m and M. The mass of M is twice that of m. The spring is released so that the masses move in opposite directions.

What is \(\frac{\text { kinetic energy of } m}{\text { kinetic energy of } M}\) ?

A. \(\frac{1}{2}\)

B. 1

C. 2

D. 4

**▶️Answer/Explanation**

Ans:C

As net force is zero , momentum will conserved on system

$p_m=p_M$

$\sqrt{2mK_m}=\sqrt{2MK_M}$

$\frac{\text { kinetic energy of } m}{\text { kinetic energy of } M}=\frac{M}{m}\Rightarrow 2$

### Question

An object of mass \(M\) is accelerated vertically upwards by a motor at a constant acceleration. The object is initially at rest and reaches a vertical speed of \(4.0 \mathrm{~ms}^{-1}\) in \(2.0 \mathrm{~s}\).

What is the average power output of the motor?

A. \(8 M\)

B. \(24 \mathrm{M}\)

C. \(32 \mathrm{M}\)

D. \(48 M\)

**▶️Answer/Explanation**

Ans:B

\(\begin{aligned} s & =\left(\frac{u+v}{2}\right) t \\ h & =\left(\frac{0+4}{2}\right) \times 2 \\ & =4 \mathrm{~m}\end{aligned}\)

\( \begin{aligned} P & =\frac{\text { total work }}{\text { total time }}\Rightarrow \frac{KE+PE}{2} \\ & =\frac{\frac{1}{2} \times m(4)^2+m g(4)}{2} \\ & =\frac{8 m+40 m}{2}=24 m\end{aligned} \)

### Question

An object is released from rest at \(X\) and slides to \(Y\). The vertical distance between \(X\) and \(Y\) is \(10 \mathrm{~m}\). During the motion, \(20 \%\) of the object’s initial gravitational potential energy is lost as friction.

What is the speed of the object at \(Y\) ?

A. \(\frac{16}{\sqrt{g}}\)

B. \(2 \sqrt{g}\)

C. \(4 \sqrt{g}\)

D. \(8 g\)

**▶️Answer/Explanation**

Ans:C

As 20 % lost 80 % of potential energy will convert into kinetic energy

\(\begin{aligned} & m g h \times 80 \%=\frac{1}{2} m v^2 \\ & v=\sqrt{2 \times g \times 10 \times 0.8}\Rightarrow4 \sqrt{g}\end{aligned}\)