IMO Practice Test — Work and Energy
12 Questions • 15 min • Olympiad level
15:00
Question 1 of 12
A force of 30 N is applied at $60^{\circ}$ to the displacement of 4 m. The work done is ($\cos60^{\circ}=0.5$)
$120\,\text{J}$
$60\,\text{J}$
$240\,\text{J}$
$0\,\text{J}$
Explanation: $W=Fs\cos\theta=30\times4\times0.5=60\,\text{J}$.
Question 2 of 12
A body of mass 2 kg has 64 J of kinetic energy. Its speed is
$4\,\text{m/s}$
$8\,\text{m/s}$
$16\,\text{m/s}$
$32\,\text{m/s}$
Explanation: $64=\frac{1}{2}\times2\times v^2\Rightarrow v^2=64\Rightarrow v=8\,\text{m/s}$.
Question 3 of 12
If the kinetic energy of a body is increased to 9 times, its speed becomes
3 times
9 times
81 times
$\sqrt{3}$ times
Explanation: Since $E_k\propto v^2$, a 9-fold energy means speed becomes $\sqrt{9}=3$ times.
Question 4 of 12
A 0.5 kg ball is dropped from 20 m. Its speed just before hitting the ground ($g=10\,\text{m/s}^2$) is
$10\,\text{m/s}$
$20\,\text{m/s}$
$40\,\text{m/s}$
$200\,\text{m/s}$
Explanation: $mgh=\frac{1}{2}mv^2\Rightarrow v^2=2gh=400\Rightarrow v=20\,\text{m/s}$.
Question 5 of 12
A 1000 kg car moving at 20 m/s is braked to rest. The work done by the brakes is
$2\times10^5\,\text{J}$
$4\times10^5\,\text{J}$
$-4\times10^5\,\text{J}$
$-2\times10^5\,\text{J}$
Explanation: $W=\Delta E_k=0-\frac{1}{2}\times1000\times20^2=-2\times10^5\,\text{J}$.
Question 6 of 12
A pump lifts 600 kg of water to 15 m in 30 s ($g=10\,\text{m/s}^2$). Its power is
$300\,\text{W}$
$3000\,\text{W}$
$9000\,\text{W}$
$90000\,\text{W}$
Explanation: $W=mgh=600\times10\times15=90000\,\text{J}$; $P=\frac{90000}{30}=3000\,\text{W}$.
Question 7 of 12
At the midpoint of a freely falling body's path, the ratio of its kinetic energy to potential energy (measured from the start) is
1 : 1
2 : 1
1 : 2
1 : 4
Explanation: At half the height, PE has halved and the lost PE became KE, so KE = PE, a 1:1 ratio.
Question 8 of 12
Two bodies of masses $m$ and $2m$ have equal kinetic energies. The ratio of their speeds (lighter : heavier) is
$1:1$
$\sqrt{2}:1$
$1:\sqrt{2}$
$2:1$
Explanation: $\frac{1}{2}mv_1^2=\frac{1}{2}(2m)v_2^2\Rightarrow v_1^2=2v_2^2\Rightarrow v_1:v_2=\sqrt{2}:1$.
Question 9 of 12
A 2 kW appliance runs for 30 minutes. The energy consumed is
$1\,\text{kWh}$
$2\,\text{kWh}$
$0.5\,\text{kWh}$
$60\,\text{kWh}$
Explanation: Energy $=P\times t=2\,\text{kW}\times0.5\,\text{h}=1\,\text{kWh}$.
Question 10 of 12
A ball thrown vertically upward momentarily stops at the top. At that instant its kinetic energy is
maximum
zero
equal to its weight
negative
Explanation: Speed is zero at the highest point, so $E_k=\frac{1}{2}mv^2=0$ (all energy is potential).
Question 11 of 12
The work done in holding a 5 kg bag stationary for 10 minutes is
$500\,\text{J}$
$3000\,\text{J}$
$0\,\text{J}$
$50\,\text{J}$
Explanation: There is no displacement, so $W=Fs=0$ regardless of time held.
Question 12 of 12
A 60 W bulb and a 100 W bulb both run for 2 hours. The extra energy used by the 100 W bulb is
$40\,\text{Wh}$
$80\,\text{Wh}$
$160\,\text{Wh}$
$200\,\text{Wh}$
Explanation: Extra power $=40\,\text{W}$; extra energy $=40\times2=80\,\text{Wh}$.