Question

A body of mass 500 g is falling from rest from a height of 3.2 m from the ground. If the body reaches the ground with a velocity of 6 ms⁻¹, then the energy lost by the body due to air resistance is (Acceleration due to gravity = 10 ms⁻²)

• A. 14 J
• B. 7 J
• C. 21 J
• D. 28 J

Hint:

An alternative approach is the Work-Energy Theorem: \(W_{total} = \Delta KE\). Here, total work is done by gravity and air resistance: \(W_g + W_{air} = KE_f - KE_i\). Calculating \(W_g = mgh = 16\) J gives \(16 + W_{air} = 9 - 0\), so \(W_{air} = -7\) J. The energy lost is the magnitude of the work done by resistance, which is 7 J.

Answer 1

The Correct Option is B

Step 1: Identify Parameters: Mass, \( m = 500 \, \text{g} = 0.5 \, \text{kg} \). Initial height, \( h = 3.2 \, \text{m} \). Initial velocity, \( u = 0 \) (rest). Final velocity at ground, \( v = 6 \, \text{ms}^{-1} \). Gravity, \( g = 10 \, \text{ms}^{-2} \).
Step 2: Calculate Initial Mechanical Energy: At the top, the energy is purely potential. \[ E_{initial} = mgh = 0.5 \times 10 \times 3.2 = 16 \, \text{J} \]
Step 3: Calculate Final Mechanical Energy: At the bottom, potential energy is zero (reference level), and energy is purely kinetic. \[ E_{final} = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.5 \times (6)^2 \] \[ E_{final} = 0.25 \times 36 = 9 \, \text{J} \]
Step 4: Calculate Energy Lost: According to the conservation of energy, the difference between initial and final mechanical energy is the work done against resistive forces (air resistance). \[ \text{Energy Lost} = E_{initial} - E_{final} \] \[ \text{Energy Lost} = 16 \, \text{J} - 9 \, \text{J} = 7 \, \text{J} \]