A block of mass 10 kg is moving in x-direction with a constant speed of 10 m/s. It is subjected to a retarding force F = -0.1 J/m during its travel from x = 20 m to x = 30 m. Its final kinetic energy will be:
Correct answer: A. 475 J
- A. 475 J
- B. 450 J
- C. 275 J
- D. 250 J
Explanation
m=10kgv=10m/sFret=-0.1J/mAt x= 20mF= -0.1 (20)=2NAt x=30mF=-0.1 (30)=3NAverage force=2+3/2 = 5/2 = 2.5NSo the average work done by the retarding force will be; = F avg × d = 2.5 × 10 = 25JTherefore, K.E final = K.E initial - work done by FretK.E final = ½ (10)(10)2 - 25K.E final = 475 JHence,A is the correct Answer.
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The work-energy theorem states that the net work done on an object equals its change in kinetic energy. Questions involve positive and negative work, variable forces and force-displacement graphs, acceleration or deceleration, and the difference between this theorem and conservation of mechanical energy, which also requires attention to potential energy and non-conservative work.
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