Online Test: Work, Energy and Power | Class 11 Physics

Question 1 of 18

Work done by a force is defined as:

$\vec{F}\times\vec{s}$

$\vec{F}\cdot\vec{s}$

$\frac{\vec{F}}{\vec{s}}$

$\vec{F}+\vec{s}$

Explanation: Work is the scalar (dot) product of force and displacement: $W=\vec{F}\cdot\vec{s}=Fs\cos\theta$.

Question 2 of 18

A force of 10 N moves a body 4 m at $60^\circ$ to the displacement. The work done is:

$20\ \text{J}$

$40\ \text{J}$

$10\ \text{J}$

$5\ \text{J}$

Explanation: $W=Fs\cos\theta=10\times 4\times\cos 60^\circ=40\times 0.5=20\ \text{J}$.

Question 3 of 18

The kinetic energy of a 2 kg body moving at 5 m/s is:

$10\ \text{J}$

$25\ \text{J}$

$50\ \text{J}$

$5\ \text{J}$

Explanation: $KE=\frac{1}{2}mv^2=\frac{1}{2}\times 2\times 5^2=25\ \text{J}$.

Question 4 of 18

The work-energy theorem states that the net work done equals:

change in momentum

change in kinetic energy

change in potential energy

total energy

Explanation: $W_{net}=\Delta KE$ — the net work done on a body equals the change in its kinetic energy.

Question 5 of 18

For a variable force, work done equals the:

slope of the F-x graph

area under the F-x graph

intercept of the F-x graph

length of the F-x curve

Explanation: $W=\int F\,dx$ is the area under the force-displacement graph.

Question 6 of 18

If the momentum of a body is doubled, its kinetic energy becomes:

double

half

four times

unchanged

Explanation: $KE=\frac{p^2}{2m}\propto p^2$; doubling $p$ makes $KE$ four times larger.

Question 7 of 18

Gravitational potential energy of a body of mass $m$ at height $h$ is:

$\frac{1}{2}mh$

$mgh$

$\frac{1}{2}mgh^2$

$mg$

Explanation: Near the Earth's surface, $U=mgh$ relative to a chosen zero level.

Question 8 of 18

The potential energy stored in a spring of constant $k$ stretched by $x$ is:

$kx$

$\frac{1}{2}kx^2$

$kx^2$

$\frac{1}{2}kx$

Explanation: Spring PE is the area under $F=kx$, giving $U=\frac{1}{2}kx^2$.

Question 9 of 18

Which of the following is a conservative force?

friction

air resistance

gravitational force

viscous force

Explanation: Gravity does path-independent work and zero work over a closed loop, so it is conservative.

Question 10 of 18

A stone dropped from 20 m hits the ground at (take $g=10\ \text{m/s}^2$):

$10\ \text{m/s}$

$15\ \text{m/s}$

$20\ \text{m/s}$

$25\ \text{m/s}$

Explanation: By energy conservation $v=\sqrt{2gh}=\sqrt{2\times 10\times 20}=\sqrt{400}=20\ \text{m/s}$.

Question 11 of 18

The SI unit of power is the:

joule

newton

watt

pascal

Explanation: Power is work per unit time, measured in watts ($1\ \text{W}=1\ \text{J/s}$).

Question 12 of 18

Instantaneous power equals:

$\frac{\vec{F}}{\vec{v}}$

$\vec{F}\cdot\vec{v}$

$\vec{F}\times\vec{v}$

$\frac{\vec{v}}{\vec{F}}$

Explanation: $P=\frac{dW}{dt}=\vec{F}\cdot\vec{v}=Fv\cos\theta$.

Question 13 of 18

A pump raises 200 kg of water 10 m in 10 s. Its power is (take $g=10\ \text{m/s}^2$):

$1000\ \text{W}$

$2000\ \text{W}$

$200\ \text{W}$

$20000\ \text{W}$

Explanation: $P=\frac{mgh}{t}=\frac{200\times 10\times 10}{10}=2000\ \text{W}$.

Question 14 of 18

In every collision the quantity always conserved is:

kinetic energy

linear momentum

speed

potential energy

Explanation: Linear momentum is conserved in all collisions; kinetic energy only in elastic ones.

Question 15 of 18

When two equal masses collide elastically in 1D and one is initially at rest, they:

stick together

both stop

exchange velocities

move with the same velocity

Explanation: For equal masses in an elastic collision, velocities are exchanged: the moving one stops and the other moves off.

Question 16 of 18

A 2 kg body at 3 m/s sticks to a 1 kg body at rest. The common velocity is:

$1\ \text{m/s}$

$2\ \text{m/s}$

$1.5\ \text{m/s}$

$3\ \text{m/s}$

Explanation: $v=\frac{2\times 3+1\times 0}{2+1}=\frac{6}{3}=2\ \text{m/s}$.

Question 17 of 18

The coefficient of restitution for a perfectly inelastic collision is:

$1$

$0.5$

$0$

between 0 and 1

Explanation: In a perfectly inelastic collision the bodies stick, so the relative velocity of separation is zero, giving $e=0$.

Question 18 of 18

A ball dropped from 4 m rebounds to 1 m. The coefficient of restitution is:

$0.25$

$0.5$

$0.75$

$1$

Explanation: $e=\sqrt{\frac{h_2}{h_1}}=\sqrt{\frac{1}{4}}=0.5$.

Online Test — Work, Energy and Power