A 2.0-kg block slides down a rough track that has a complicated curved shape. Th

Question

A 2.0-kg block slides down a rough track that has a complicated curved shape. The block has a speed of 5.0 m/s after its height above the starting point has decreased by 1.8 m. Assume the block starts from rest. How much work is done on the block by the force of friction during this descent?[Hint: Use work and energy, not Newton's laws of motion. The latter would be too complicated as the normal reaction, and therefore the friction force, is constantly changing in a complicated way during the descent.]

Explanation / Answer

Data: Mass, m = 2 kg Displacement, s = 1.8 m Final velocity, vf = 5 m/s Initial velocity, vi = 0 m/s Let F = Gravitational force    = m g    = 2 * 9.8    = 19.6 N W'= Work done byFriction Solution: From work-energy theorem, Total work done = Change in kinetic energy F * s + W' = (1/2) m ( vf^2 - vi^2 ) ( 19.6 * 1.8 ) + W'= 0.5 * 2 * ( 5^2 - 0 ) 35.28 + W'= 25 W' = 25 - 35.28       = - 10.28 J       ˜ - 10 J Ans: Work done by friction = - 10 J

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