Question

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.}

• A. 1980
• B. 2020
• C. 2000
• D. 1000

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The ball starts from rest, falls, penetrates the sand, and comes to rest again. Thus, its total change in kinetic energy is zero.
By the Work-Energy Theorem, the net work done by all forces on the ball must equal zero.
The two forces acting on the ball are gravity (acting over the entire fall and penetration distance) and the resisting force of the sand (acting only during penetration).
Step 2: Key Formula or Approach:
Work-Energy Theorem: \(W_{net} = \Delta K = 0\).
\(W_{gravity} + W_{sand} = 0\).
\(W_{gravity} = mg(h + d)\), where \(h\) is the fall height and \(d\) is the penetration depth.
\(W_{sand} = -F_{avg} \cdot d\), where \(F_{avg}\) is the upward average force exerted by the sand.
Step 3: Detailed Explanation:
Given values:
Mass \(m = 2 \text{ kg}\)
Height \(h = 10 \text{ m}\)
Penetration depth \(d = 10 \text{ cm} = 0.1 \text{ m}\)
Gravity \(g = 10 \text{ m/s}^2\)
Calculate the total work done by gravity:
\[ W_{gravity} = mg(h + d) = 2 \times 10 \times (10 + 0.1) \] \[ W_{gravity} = 20 \times 10.1 = 202 \text{ J} \] Calculate the work done by the sand:
\[ W_{sand} = -F_{avg} \times 0.1 \] Apply the Work-Energy Theorem:
\[ 202 - 0.1 F_{avg} = 0 \] \[ 0.1 F_{avg} = 202 \implies F_{avg} = \frac{202}{0.1} = 2020 \text{ N} \] Step 4: Final Answer:
The average force exerted by the sand is \(2020 \text{ N}\).