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/s^{2} in the first 15 s, what is the frictional force between the surfaces?

Resultant force on box = 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 a

a = – 2 m/s^{2}

The velocity of box at time 15s is required.

Hence when t = 0s, box accelerates at 2m/s^{2} 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) / 5

v = 20 m/s. ** **

**Therefore, velocity decreases at a constant rate of -2 m/s ^{2} 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.

October 25, 2010 at 5:57 PM

Opps sorry. The acceleration is missing in the question. Just amended it. So it is given that the acceleration is 2 m/s^2 for first 15s. Thanks Ying Chieh. =)