A factory worker pushes a 298 kg crate a distance of 43 m al

A factory worker pushes a 29.8 kg crate a distance of 4.3 m along a level floor at constant velocity by pushing downward at an angle of 32 below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.24.

Part A

What magnitude of force must the worker apply to move the crate at constant velocity?

Part B

How much work is done on the crate by this force when the crate is pushed a distance of 4.3 m ?

Part C

How much work is done on the crate by friction during this displacement?

Part D

How much work is done by the normal force?

Part E

How much work is done by gravity?

Part F

What is the total work done on the crate?

Solution

Let F is the required force.

A) Normal force acting on the crate, N = m*g + F*sin(32)

when the crate moves with constant net force on crate is zero.

Fnet = 0

F*cos(32) - N*mue_k = 0

F*cos(32) - (m*g + F*sin(32))*mue_k = 0

F*(cos(32) - mue_k*sin(32)) = mue_k*m*g

F = mue_k*m*g/(cos(32) - mue_k*sin(32))

= 0.24*29.8*9.8/(cos(32) - 0.24*sin(32))

= 97.23 N
b)

Workdone by applied force = F*d*cos(32)

= 97.23*4.3*cos(32)

= 354.56 J

c)

Workdone by friction = mue_k*N*d*cos(180)

= mue_k*(m*g + F*sin(32))*d*(-1)

= 0.24*(29.8*9.8 + 97.23*sin(32))*4.3*(-1)

= -354.56 J

d)

Workdone by nrmal force = N*d*cos(90)

= 0

e) Workdone by gravity = m*g*d*cos(90)

= 0

f) Total workdone = 354.56 - 354.56 + 0 + 0

= 0