A 2.2 kg block starts from rest on a rough inclined plane that makes an angle of 25 degree with the horizontal. the coefficient of kinetic friction is 0.25 . as the block goes 2 m down the plane the mechanical energy of the earth block system chenges by ?
Answers
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
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: