An object of mass 20.0 kg is at rest at the top of an inclined plane whose length is 8.00 m and which makes an angle of 30.0° with the horizontal. The coefficient of kinetic friction between the object and the inclined plane is 0.400. The object is released and slides down the plane. What is the kinetic energy of this object at the bottom of the inclined plane? Use g = 10.0 m/s².

Question

Quizplus · Accepted Answer

Explanation:The work done by gravity is:$$W_g = mgh = (20.0 	ext{ kg})(10.0 	ext{ m/s}^2)(8.00 	ext{ m}) = 1600 	ext{ J}$$The work done by friction is:$$W_f = -f_kd = -(0.400)(20.0 	ext{ kg})(10.0 	ext{ m/s}^2)(8.00 	ext{ m}) = -640 	ext{ J}$$The net work done on the object is:$$W_{net} = W_g + W_f = 1600 	ext{ J} - 640 	ext{ J} = 960 	ext{ J}$$The kinetic energy of the object at the bottom of the inclined plane is:$$K = W_{net} = 960 	ext{ J}$$Therefore, the correct answer is A: 246 J.

An Object of Mass 20