A box, with a weight of mg = 25 N, is placed at the top of a ramp and released f

Question

A box, with a weight of mg = 25 N, is placed at the top of a ramp and released from rest. The ramp measures 4.50 meters horizontally and 3.00 meters vertically. The box accelerates down the incline, attaining a kinetic energy at the bottom of the ramp of 40.0 J. There is a force of kinetic friction acting on the box as it slides down the incline.

(a) How much work does the normal force do on the box as the box slides down the incline? (J)

(b) Calculate the change in gravitational potential energy that the box experiences in this process. (Use the appropriate sign.) (J)

(c) How much work does the force of friction do on the box as the box slides down the incline? (Use the appropriate sign.) (J)

(d) What is the coefficient of kinetic friction between the box and ramp?

Explanation / Answer

(a) work done by the normal force = 0 J (since force and displacement are perpendicular)

(b) change in potential energy = - (25*3.3) = - 82.5 J

(c) work done by the friction = 42 - 82.5 = - 40.5 J

(d) friction force f = 40.5/sqrt(4.5^2+3.3^2) = 7.258 N
but f = *25*cos where cos = 4.5/5.58 = 0.8064

*25*cos = 7.258

= 0.36

so,the coefficient of kinetic friction = 0.36

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