JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Laws of Motion

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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:

(A)40 N
(B)5 N✓ Correct
(C)20 N
(D)10 N

Answer: (B)

Solution

Once the force is removed the body moves at constant velocity, covering

50

m in

10

s, so that velocity is

v = \frac{50}{10} = 5

m/s. This speed was reached from rest during the

20

s the force acted, so

v = a t

gives

a = \frac{5}{20} = 0.25

m/s

^{2}

. The force is

F = ma = 20 \times 0.25 = 5

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Single correct option

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