When velocity of body of mass 10 kg is increased from 2ms⁻¹ to 8 ms⁻¹ then the work done on the body is:
Correct answer: C. 300 J
- A. 100 J
- B. 200 J
- C. 300 J
- D. 400 J
Explanation
ΔKE = ½ m (v₂² − v₁²) = 0.5×10×(64−4) = 5×60 = 300 J. Thus option C.
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Work transfers energy when a force causes displacement, while kinetic energy, gravitational potential energy and power describe motion, position and the rate of energy transfer. The work-energy theorem links net work with change in kinetic energy, and efficiency accounts for energy losses rather than treating them as energy destruction.
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