JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2026
- Medium
- MCQ
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/s\(^2\))
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(2020\)
Step-by-step Solution
Detailed explanation
Using the work-energy theorem, the total work done on the ball is equal to its change in kinetic energy.
Since the ball starts from rest and finally comes to rest, the change in kinetic energy is zero.
\(W_{\text{gravity}} + W_{\text{sand}} = \Delta K = 0\)
The work done by gravity is over the total distance \(h + d\), and the work done by the resistive force of the sand \(F\) is over the distance \(d\).
\(mg(h + d) - Fd = 0\)
\(F = \dfrac{mg(h + d)}{d}\)
Given \(m = 2\) kg, \(g = 10\) m/s\(^2\), \(h = 10\) m, and \(d = 10\) cm \(= 0.1\) m:
\(F = \dfrac{2 \times 10 \times (10 + 0.1)}{0.1}\)
\(F = \dfrac{20 \times 10.1}{0.1}\)
\(F = 20 \times 101 = 2020\) N
Answer: \(2020\)
Since the ball starts from rest and finally comes to rest, the change in kinetic energy is zero.
\(W_{\text{gravity}} + W_{\text{sand}} = \Delta K = 0\)
The work done by gravity is over the total distance \(h + d\), and the work done by the resistive force of the sand \(F\) is over the distance \(d\).
\(mg(h + d) - Fd = 0\)
\(F = \dfrac{mg(h + d)}{d}\)
Given \(m = 2\) kg, \(g = 10\) m/s\(^2\), \(h = 10\) m, and \(d = 10\) cm \(= 0.1\) m:
\(F = \dfrac{2 \times 10 \times (10 + 0.1)}{0.1}\)
\(F = \dfrac{20 \times 10.1}{0.1}\)
\(F = 20 \times 101 = 2020\) N
Answer: \(2020\)
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