Question

A force of 4N acts on a body of mass 2 kg for 4 seconds. Assuming the body to be initially at rest and find the distance covered in 10 seconds after force starts acting on it

Answer 1

Force = 4 N., Mass = 2 kg.

Using the Formula,

F = ma , a = F/m    a = 4/2

a = 2 m/s²

Now, u = 0

t = 4 seconds

v - u = at

⇒ v - 0 = 2 × 4

⇒ v = 8 m/s.

Now, v² - u² = 2aS

64 = 2 × 2 × S

S = 16 m.

It is the distance covered in first 4 seconds.

Now, for the distance covered in another 6 seconds.

u =  v of above = 8 m/s.

t = 10 - 4 = 6 seconds.

a = 0 [Force is not acting now]

S = ut + 1/2at²

∴ S = 8 × 6 +  0

∴ S = 48 m.

Total distance covered in 10 seconds = 16 + 48 = 64 m.

Answer 2

A force of 4N acts on a body of mass 2kg 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 × 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 × 4 + 1/2 × 2 × 4²} + {8 × 6 + 1/2 × 0 × 6²}
= 16 + 48 = 64m