A 2.2 kg block starts from rest on a rough inclined plane that makes an angle of 25° with the horizontal.
The coefficient of kinetic friction is 0.25. As the block goes 2.0 m down the plane, the mechanical energy
of the Earth-block system changes by
(1) Zero
(2) -9.8 J
(3) 9.8 J
(4) – 18 J
(5) 18 J
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Answer:
Mechanical energy is considered as the sum of KE & PE of the body.
The plant was considered frictionless and no energy will be lost through it like sound energy.
The force due to friction = mu * N , where "mu" is the coefficient of friction, and N is the normal force.
Work done by friction = Fd = (mu * N) * d = mu * mgcosθ * d. Hence the work done = 0.25 * (2.2)(9.8)cos25° * 2 = 9.8 Nm
Explanation:
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