Question

A 17.5-kg block is dragged over a rough, horizontal surface by a 75-N force acting at 21° above the horizontal. The block is displaced 5.7 m, and the coefficient of kinetic friction is 0.150.
Requried:
a. Find the work done on the block by the 75-N force.
b. Find the work done on the block by the normal force.
c. Find the work done on the block by the gravitational force.
d. What is the increase in internal energy of the block-surface system due to friction?
e. Find the total change in the block's kinetic energy.

Answer 1

Answer:

Explanation:

given

m= 17.5kg

F= 75N

d= 5.7m

∪=0.150

θ= 21°

a. W = Fcos θ × d

75cos21° ×5.7

=399.106J

b. normal force is zero. 0 Joules

cos 90°=0