Question

a body of mass 1 kg is moving with a momentum of 1 N s for 2 sec after this a force of 5 N starts acting on it for 2 sec the change in K. E
occurs will be​

Answer 1

Explanation:

Given -

• A body of mass 1 Kg is moving with a momentum of 1 N.s
• Force of 5 N starts acting on it for 2 sec

To Find -

• Change in K.E occur will be ?

Solution -

As we know that :-

• P = mv

1 = 1 v

v = 1 m/s

As F = 5N

Then, F = ma

5 = 1 a

a = 5 m/s²

At t = 0,

$\mathsf{K.E_{i} = \dfrac{1}{2} m {v}^{2} }$

$\mathsf{ \dfrac{1}{2} \times 1 \times {(1)}^{2} }$

$\mathsf{K.E_{i} = 0.5J}$

Now, At t = 2 s

$\mathsf{v_{f} = u + at}$

$\mathsf{v_{f} = 1 + 5 \times 2}$

• $\mathsf{v_{f} = 11ms {}^{ - 1} }$

Therefore,

$\mathsf{K.E_{f} = \dfrac{1}{2} m {v}^{2} }$

$\mathsf{ \dfrac{1}{2} \times 1 \times {(11)}^{2} }$

$\mathsf{\frac{121}{2} }$

$\mathsf{K.E_{f} = 60.5J}$

Then,

$\mathsf{ \triangle{K.E} = K.E_{f} - K.E_{i} }$

→ 60.5J - 0.5J

→ 60J

Hence,

The change in K.E occurs will be 60J.