To push a 25.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel to the incline. As the crate slides 1.7 m, how much work is done on the crate by (a) the worker's applied force, (b) the gravitational force on the crate, and (c) the normal force exerted by the incline on the crate? (d) What is the total work done on the crate?

Answer

given,

Mass = 25 Kg

angle inclined with horizontal = 25.0°

force exerted is equal to = 209 N

a) work done = F. s

work done = 209 * 1.7

work done = 355.3 J

b) gravitational force is acting downward

component along the incline is = mg sin (θ)

= mg sin 25°

but is acting opposite to the motion of the crate so, work done by this will be negative.

work = - mg sin 25°*1.7

= - 176.2 J

c) the normal force exerted by

the incline on the crate = mg cos (25°)

=222.045 J

d) total work

= 355.3 - 176.2

= 179.1 J