Force acts for $$20$$ s on a body of mass $$20$$ kg, starting from rest, after which the force ceases and then body describes $$50$$ m in the next $$10$$ s. The value of force will be:

Solution :

Given :

Mass of body,

$$m = 20\text{ kg}$$

Force acts for,

$$t_1 = 20\text{ s}$$

Distance travelled in next 10s after force ceases :

$$s = 50\text{ m}$$

After the force ceases, body moves with constant velocity.

Therefore,

$$v = \frac{s}{t}$$

$$= \frac{50}{10}$$

$$= 5\text{ m s}^{-1}$$

This is the velocity acquired after 20s.

Using,

$$v = u + at$$

Since body starts from rest,

$$u = 0$$

$$5 = 0 + a(20)$$

$$a = \frac{5}{20}$$

$$= 0.25\text{ m s}^{-2}$$

Using Newton’s second law :

$$F = ma$$

$$= 20 \times 0.25$$

$$= 5\text{ N}$$

Final Answer :

$$5\text{ N}$$