Physics·MCQ Practice

Work — MCQ Practice

NEET UG
Version 1Updated 22 Mar 2026

Interactive MCQ Practice

Test your knowledge. Click “Solve” to reveal options, select your answer, then check the result. 7 questions available.

Q1medium

A block of mass 2 kg2\text{ kg} is pulled 5 m5\text{ m} along a horizontal surface by a force of 10 N10\text{ N} acting at an angle of 3737^\circ above the horizontal. The coefficient of kinetic friction between the block and the surface is 0.20.2. Calculate the net work done on the block. (Given: sin37=0.6\sin 37^\circ = 0.6, cos37=0.8\cos 37^\circ = 0.8, g=10 m/s2g = 10\text{ m/s}^2)

Q2medium

A particle moves along the x-axis from x=0x = 0 to x=4 mx = 4\text{ m} under the influence of a force F(x)=(5x22x+3) NF(x) = (5x^2 - 2x + 3)\text{ N}. Calculate the work done by the force.

Q3medium

A 0.5 kg0.5\text{ kg} ball is thrown vertically upwards with an initial speed of 10 m/s10\text{ m/s}. Calculate the work done by gravity on the ball when it reaches its maximum height. (Take g=10 m/s2g = 10\text{ m/s}^2)

Q4easy

A spring with a spring constant k=200 N/mk = 200\text{ N/m} is compressed by 10 cm10\text{ cm} from its equilibrium position. Calculate the work done by the spring during this compression.

Q5medium

A 10 kg10\text{ kg} object is initially at rest. A constant net force of 50 N50\text{ N} acts on it for 4 s4\text{ s}. What is the work done by the net force on the object?

Q6medium

A force-displacement graph for a particle is shown below. The force acts along the direction of displacement. Calculate the work done by the force from x=0 mx = 0\text{ m} to x=6 mx = 6\text{ m}. (Imagine a graph: x-axis from 0 to 6, y-axis (Force in N) from 0 to 10. The graph is a straight line from (0,0) to (2,10), then a horizontal line from (2,10) to (4,10), then a straight line from (4,10) to (6,0).)

Q7medium

A 5 kg5\text{ kg} object is lifted vertically upwards by a rope with a constant acceleration of 2 m/s22\text{ m/s}^2. If the object is lifted by 3 m3\text{ m}, what is the work done by the tension in the rope? (Take g=10 m/s2g = 10\text{ m/s}^2)

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