Question

a factory worker pushes a crate of mass 28.2kg a distance of 4.70m aong a level floor at constant velocity by pushing horizontally on it. the coefficient of kinetic friction between the crate and the floor is .24.

A) what magnitude of the force must the wprker apply?

B) how much work is done on the crate by this force?

c)how much work is done on the crate by friction??

d)how much work is done by the normal force?

e) how much work is done by gravity?

f)what is the total work done on the crate?

Answer 1

part A:

as it travels with constant velocity, net force =0

magnitude of the force to be applied-friction force=0

==>magnitude of he force to be applied=friction force=fiction coefficient*normal force

=0.24*mass*acceleration due to gravity

=0.24*28.2*9.8

=66.326 N

part B:

work done on the crate by this force

=force*distance

=66.326*4.7

=311.73 J

part C:

friction is acting in opposite direction of the direction of motion.

work done on the crate by friction=friction force*distance*cos(180)

=66.326*4.7*(-1)

=-311.73 J

part d:

normal force is perpendicular to the motion direction

as cos(90)=0

work done by normal force=force*distance*cos(90)=0

part e:

as gravity is perpendicular to the direction of motion, work done=0

part f:

total work done=311.73-311.73=0 J