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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