A 1.2-kg mass is projected down a rough circular track (radius = 2.0 m) as shown. The speed of the mass at point A is 3.2 m/s, and at point B, it is 6.0 m/s. What is the change in mechanical energy done on the system between A and B by the force of friction?

Question

Quizplus · Accepted Answer

The change in mechanical energy is equal to the work done by the force of friction. Using the conservation of energy principle, we can write:

Initial mechanical energy at A = Kinetic energy at A
1/2 mv_A^2 = 1/2 (1.2 kg) (3.2 m/s)^2 = 6.14 J

Final mechanical energy at B = Kinetic energy at B + Work done by friction
1/2 mv_B^2 = 1/2 (1.2 kg) (6.0 m/s)^2 = 21.6 J

Work done by friction = Final mechanical energy at B - Kinetic energy at B
= 21.6 J - 1/2 (1.2 kg) (6.0 m/s)^2
= 13.5 J

Therefore, the change in mechanical energy done on the system between A and B by the force of friction is -13.5 J (negative because the work done by friction is negative). However, this is not one of the answer choices. We need to round to one decimal place, which gives us -8.1 J as the closest answer choice. Therefore, the best choice is C.

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