Question

8. Il An 8.00 kg package in a mail-sorting room slides 2.00 m down a

chute that is inclined at 53.0° below the horizontal. The coefficient o
kinetic friction between the package and the chute's surface is 0.40.
Calculate the work done on the package by (a) friction, (b) gravity, and
(c) the normal force. (d) What is the net work done on the package?

Answer 1

See below

Step-by-step explanation:

Normal force = m g cos 53 = 8 kg * 9.8 m/s^2 * cos 53 = 47.1823 N

no work is done by this force

Force friction = coeff friction * force normal = .4 * 47.1823 = 7.55 N

work of friction = 7.55 * 2 m = 15.1 j

Force Downplane = mg sin 53 = 62.61 N

work = 62.61 * 2 = 125.22 j

Net Force downplane = force downplane - force friction = 55.06 N

net Work = force * distance = 55.06 N * 2 M = 110.12 j