Question

12. A force of 4 N acts on a body of mass 2 kg for 4 s. Assuming the body to be initially at rest find a. its velocity
when the force stops acting b. the distance covered in 10 s after the force starts acting.​

Answer 1

Answer:

64 m

Explanation:

✔️A force of 4N acts on a body of mass 2 kg. for 4 seconds.

✔️mass of body, m = 2kg

✔️initial velocity of body , u = 0 m/s force acts on the body , F = 4N

from Newton's 2nd law⤵️

F = ma

4 = 2a => a = 2m/s2

force acting on body for 4sec. so, velocity of body changes in 4 sec. use formula, v u + at

v = 0 + 2 x 4 = 8 m/s

✔️hence, body moves with velocity 8m/s . now, applied force is removed after 4sec then, body moves with constant velocity. hence, acceleration = 0

✔️we have to find distance covered in 10 sec after force starts acting on it.

✔️so, total distance = distance covered in first 4 sec + distance covered in rest 6 sec

➪ (0 x 4 + 1/2 x 2 x 4²) + (8 x 6 + 1/2 x 0 x 6²)

➪ 16 + 48 = 64m

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