Question:
The velocity-time graph of a body of mass 4 kg moving along a straight line is shown in figure. Work done by all the forces acting on the body from \( t=0 \) to \( t=5s \) is
The velocity-time graph of a body of mass 4 kg moving along a straight line is shown in figure. Work done by all the forces acting on the body from \( t=0 \) to \( t=5s \) is
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A decreasing velocity trend automatically indicates that the net work done on the system must be negative, as the body is losing kinetic energy due to dissipative braking forces.
Updated On: Jun 8, 2026
- 300 J
- -300 J
- -600 J
- 600 J
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The Correct Option is C
Solution and Explanation
Concept:
According to the Work-Energy Theorem, the net work done by all the forces acting on a body is equal to the change in its kinetic energy:
\[
W_{\text{net}} = \Delta K.E. = \frac{1}{2}m v_f^2 - \frac{1}{2}m v_i^2
\]
Step 1: Extracting initial and final velocities from the graph data.
From the velocity-time graph axes:
• Mass of the body \( m = 4 \, \text{kg} \)
• At \( t = 0 \, \text{s} \), initial velocity \( v_i = 20 \, ms^{-1} \)
• At \( t = 5 \, \text{s} \), final velocity \( v_f = 10 \, ms^{-1} \)
Step 2: Calculating the net work done.
\[ W = \frac{1}{2} (4) \left[ 10^2 - 20^2 \right] \] \[ W = 2 \cdot [100 - 400] = 2 \cdot (-300) = -600 \, \text{J} \]
Step 1: Extracting initial and final velocities from the graph data.
From the velocity-time graph axes:
• Mass of the body \( m = 4 \, \text{kg} \)
• At \( t = 0 \, \text{s} \), initial velocity \( v_i = 20 \, ms^{-1} \)
• At \( t = 5 \, \text{s} \), final velocity \( v_f = 10 \, ms^{-1} \)
Step 2: Calculating the net work done.
\[ W = \frac{1}{2} (4) \left[ 10^2 - 20^2 \right] \] \[ W = 2 \cdot [100 - 400] = 2 \cdot (-300) = -600 \, \text{J} \]
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