A 10 kg box is initially at rest on a rough horizontal surface and a horizontal force is then applied to it. The force varies as shown in the diagram.

a) If the box moves with an acceleration of 2m/s2 in the first 15 s, what is the frictional force between the surfaces?

Resultant force on box F = ma

Applied force – frictional force = ma

40 – frictional force = 10 x 2

** ** frictional force = 20 N

b) How does the velocity change during the next 5 s?

Next 5 s after time 15 s, there is no applied force. Hence the resultant force acting on box isonly frictional force 20N (in direction opposite to motion of box).

F = ma-20 = 10 x aa = – 2 m/s2

The velocity of box at time 15s is required.

Hence when t = 0s, box accelerates at 2m/s2 from rest,

a = (v – u) / t

2 = (v – 0) /15

v = 30 m/s

(Hence the final velocity of the box at time 15s is 30 m/s, just before applied force of 40 N is released)

Find velocity of box at end of next 5s (time = 20s):

a = (v – u) /t– 2 = (v – 30) / 5v = 20 m/s. * *

*Therefore, velocity decreases at a constant rate of -2 m/s2 from 30 m/s to 20 m/s in next 5s.*

c) Plot a velocity-time graph and calculate the distance travelled by the box in the 30 s.

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