To pull a 53 kg crate across a horizontal frictionless floor, a worker applies a force of 180 n, directed 35° above the horizontal. as the crate moves 2.9 m, what work is done on the crate by (a) the worker's force, (b) the gravitational force on the crate, and (c) the normal force on the crate from the floor? (d) what is the total work done on the crate?

Answer

given,

mass of the crate  = 53 Kg

force applied by the worker = 180 N

Angle made with the horizontal = 35°

crate moves = 2.9 m

a) The work done equals the force in the direction of the displacement, times the displacement

W = F_x ×d

W = 180 cos 35° × 2.9

W = 427.6 J

b) A force that is perpendicular to the direction of the displacement does not do any work so work done by gravitational force is zero

c) Similarly for the normal force the work done will be zero.

d) Total work done by the crate is equal to 427.6 J

this is a good question! i'll do some research and come back when i have the answer!