Question

Q.6 A force of 6.0 N acts on a body of mass 3.0 kg for 7.0 s. Assuming the body to be initially at rest, find
(a) the velocity when the force stops acting,
(b) the distance covered in 12 s after the force start acting.
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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

Explanation:

a) using equation v=u+at

v= 7+6*3

v=25m/s

remember a=g