Question

The velocity-time graph for a body of mass \( 10\,\text{kg} \) is shown. Work done on the body in the first two seconds of motion is 

• A. \( -9300\,\text{J} \)
• B. \( 12000\,\text{J} \)
• C. \( -4500\,\text{J} \)
• D. \( -12000\,\text{J} \)

Hint:

Work-energy:
Use KE change when velocity known.

Answer 1

The Correct Option is C

Concept: Work done = change in kinetic energy: \[ W = \frac12 m(v^2 - u^2) \] Step 1: {\color{red}From graph.} Velocity decreases linearly from \( 50\,\text{m/s} \) to 0 in 10 s. Slope: \[ a = -5\,\text{m/s}^2 \] Step 2: {\color{red}Velocity at 2 s.} \[ v = 50 - 5\times2 = 40\,\text{m/s} \] Step 3: {\color{red}Work done.} \[ W = \frac12 \cdot 10 (40^2 - 50^2) \] \[ W = 5(1600 - 2500) = -4500\,\text{J} \]