A 40kg box on a horizontal friction surface is pulled in the horizontal direction with a force of 200N...

Question

A 40kg box on a horizontal friction surface is pulled in the horizontal direction with a force of 200N. The coefficient of kinetic friction between the surface and the box is μk=0.2. The initial velocity of the box is in the same direction with the applied horizontal force and has a magnitude of 2m/s. The box moves 6m on a straight line (g=10m/s2 ).

a) Calculate separately the work done on the object by the horizontal force and the frictional force during its movement.

b) Find the final velocity of the object using work-energy theorem.

c) Find the instantaneous power provided by the horizontal force at the last moment of the object's motion (i.e. at the end of the object's 6m path).

d) Calculate the average power provided by the horizontal force throughout the body's motion.

Answer 1

The box's weight = f_g=mg = 40x10 = 400N

The frictional force =µ x f_g = 0.2 x 400 = 80N

a) Work done by external force = f x d = 200 x 6 = 1200 J

Work done by friction = f x d = -80 x 6 = -480J

b) final Ke = initial KE + Net work done

final ke = 1/2mv² + 1200 - 480 = 0.5 x 40 x 2² +720 = 800 J

1/2mv² = 800 so 0.5 x 40 x v² = 800

v² = 40 v = 6.32 m/s

c) P = Fv = 200 x 6.32 = 1264 W

d) Average power = work done / time = 200 x 6/t

Distance d = average speed x time so t = distance/average speed = 6/(2 +6.32)/2 = 12/8.32 = 1.44 s

P = 1200/1.44 = 833W