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 4.80 meters horizontally and 3.40 meters vertically. The box accelerates down the incline, attaining a kinetic energy at the bottom of the ramp of 58.0 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?

Explanation / Answer

a) as box slides down , normal force is always in perpendicular direction to motion of box and perpendicular to displacement so always work done by normal force is zero.
w = f.d =fdcos
= 90 degrees so work done w = 0 J

b) distance travel by box in vertical direction d = 3 m
so change is mgh - mgh1
U = mgh = 25 X 3 = 75 J

c) change in energy = work done
1/2 mv2 = mgh - N.d
N = mgcos
40 = 75 - work done by friction
work done by friction = 75-40 = 35 J

d) 35 = X mgcos X d   

we get = 0.374

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