Question

27. A force of 4 newton acts on a
body of mass 2 kg for 4 second.
Assuming the body to be initially at
rest ,find (a) its velocity when the
force stop acting,(b) the distance
covered in 6 seconds after the force
is starts acting.​

Answer 1

Answer:

Explanation:

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.

hope it helps you

Answer 2

Explanation:

F=ma

4=2×a

a=2m/s^2

v=u+at

As it is at rest initials velocity is 0

Therefore,v=0+(2)(6)

v=12