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

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 4.20 meters horizontally and 3.40 meters vertically. The box accelerates down the incline, attaining a kinetic energy at the bottom of the ramp of 54.0 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? (b) Calculate the change in gravitational potential energy that the box experiences in this process. (Use the appropriate sign.) (c) How much work does the force of friction do on the box as the box slides down the incline? (Use the appropriate sign.) (d) What is the coefficient of kinetic friction between the box and ramp? Additional Materials eBook Submit Answer Save Progress

Explanation / Answer

(a) The work done by the normal force would be 0 J. (dot product gives zero when the angle is 90)

(b) mgh= -25 x 3.40 = - 85 Joules

(c) Work done by the frictional force is 54 - 85 Joules = -31 Joules

(d) Frictional force = 31/sqrt(4.22+3.42) = 5.695 N

f=u x 25 x cos(theta)

cos(theta)=4.2/5.40 = 0.777

u x 25 x 0.777 = 5.695

u=0.293.

which is the coefficient of friction.

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