Question

A horizontal force of 5N is applied on a stationary
body of mass 5kg kept on a frictionless table. The
change in K.E. of the body in 10 s is​

Answer 1

it is Correct or incorrect ,

Answer 2

Given:

A horizontal force of 5N is applied on a stationary

body of mass 5kg kept on a frictionless table.

To find:

Change in K.E. of the body in 10 sec ?

Solution:

Initial velocity = 0 m/s.

Initial kinetic energy will be :

$\therefore \: KE1 = \dfrac{1}{2} m {u}^{2}$

$\implies\: KE1 = \dfrac{1}{2} m {(0)}^{2}$

$\implies\: KE1 = 0 \: joule$

Now, let's calculate the final velocity :

$\therefore \: v = u + at$

$\implies \: v = 0 +( \dfrac{force}{mass}) t$

$\implies \: v = 0 +( \dfrac{5}{5}) \times 10$

$\implies \: v = 10 \: m {s}^{ - 1}$

So, final kinetic energy will be :

$\therefore \: KE2 = \dfrac{1}{2} m {v}^{2}$

$\implies\: KE2 = \dfrac{1}{2} \times 5 \times {(10)}^{2}$

$\implies\: KE2 =250 \: joules$

So, change in KE:

∆KE = KE2 - KE1 = 250 - 0 = 250 Joules.

So, change in KE is 250 Joules.