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

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

## 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. $$2.0 \mathrm{~m}$$

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. $18$

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

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. $$48 M$$

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. $$8 g$$

\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}