A 1.75 m tall weight-lifter raises a weight of mass 50 kg to a height of 0.5 m above its head. How much work is being done by it? (g = 10 ms-2)
Correct answer: D. 1125 J
- A. 2125 J
- B. 2500 J
- C. 50 J
- D. 1125 J
Explanation
Work done = Force x parallel distance Force = mg : 50kg x 10 N/kg = 500 NDistance = 1.75m + 0.5m = 2.25 mW.D= 2.25m x 500 N = 1125 J
This question appeared on the PMC National MDCAT 2020 paper, which you can sit online with every answer explained.
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About Work
Work is done when a force produces displacement, with its value determined by the component of force along the displacement. The topic covers positive, negative and zero work, work by variable forces from force-displacement graphs, and the connection between net work and change in kinetic energy.
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