Question

A 20 kg block is pushed 40m up an inclined plane that makes an angle of 30 degrees with the horizontal by a constant force of 250 N acting parallel to the plane. The coefficient of kinetic friction between the block and the plane is 0.30.
A)How much work is done by the 200N force?
B)What is the increase in the potential energy of the block?
C)What is the work done against friction?
D)What is the increase in the kinetic energy of the block?
E)What is the velocity of the block at the top of the incline?

Answer 1

The question seems to indicate 250N force up the incline but then also 200N. I'll go with the latter, as in your diagram. Also will us g=10m/s^2 not 9.8.

a. Work up the incline is (force)(distance) = (200N)(40m) = 8000J.

b. potential energy = m g h = 20 (10) 40 sin(30) = 4000J

c. friction force is 0.3 (normal force) = 0.3 (m g) cos(30)=52N
work against friction is (force)(distance) = (52N)(40m) = 2080J.

d. Kinetic energy (KE) is work - change in PE - friction
KE = 8000 - 4000 - 2080 = 1920J

e. KE = (1/2) m v^2
1920 = 0.5 (20kg)(v^2)
v = sqrt[2 x 1920 /20] = 13.8 m/s.