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Physics Question 36 – JEE-MAIN 2026

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A spherical ball of mass 2 kg falls from a height of 10 m and is brought to rest after penetrating 10 cm into sand. The average force exerted by sand on the ball is _______ N. (Take g=10 m/s2)

The problem can be solved by considering the total change in mechanical energy from the initial state to the final state.

Step 1: Identify Initial and Final States and Energy Changes✦ Active

The ball starts from rest at a height h=10 m and comes to rest after penetrating d=10 cm=0.1 m into the sand. The total vertical displacement from the initial position to the final resting position is Htotal=h+d. Since the ball starts and ends at rest, the total change in kinetic energy is zero.

m=2 kg Initial height h=10 m Penetration depth d=0.1 m Acceleration due to gravity g=10 m/s2
Step 2: Apply the Work-Energy Theorem○ Expand

According to the Work-Energy Theorem, the net work done on the ball is equal to its change in kinetic energy. Since the initial and final kinetic energies are zero, the net work done is zero. The forces doing work are gravity and the average resistive force from the sand.

Wnet=ΔKE=0 Work done by gravity Wg=mg(h+d) Work done by sand's force Wsand=−Favgd Therefore, Wg+Wsand=0 mg(h+d)−Favgd=0
💡 Teacher's Secret Hint

Remember that the work done by the sand's force is negative as it opposes the motion.

Step 3: Calculate the Average Force Exerted by Sand○ Expand

Rearrange the equation from Step 2 to solve for the average force Favg.

Favgd=mg(h+d) Favg=mg(h+d)d=mg(hd+1) Substitute the given values: Favg=2×10(100.1+1) Favg=20(100+1) Favg=20×101 Favg=2020 N
💡 Teacher's Secret Hint

Ensure all units are consistent (e.g., convert cm to m) before calculation.

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