JEE MAIN - Physics (2023 - 29th January Evening Shift - No. 2)

A 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:
40 N
20 N
5 N
10 N

Explanation

$$m = 20$$ kg

$$t = 20$$ sec.

Acceleration $$ = {F \over {20}}$$ m/s$$^2$$

$$\therefore$$ $$v = u + at$$

$$v = 0 + \left( {{F \over {20}}} \right)(20)$$

$$ = F$$ ms$$^{-1}$$

Now for next 10 sec.

$$S=ut$$

$$50=F(10)$$

$$F=5$$

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