Question

a body of mass 5 kg initially at rest is subjected to a force of 20 Newton what is the kinetic energy acquired by the body at the end of 10 seconds ​

Answer 1

Answer:

the final answer will e 4000 j

thus m = 5

f = 20

Answer 2

Given:

• Mass of the body, m = 5 kg

• Initial velocity, u = 0

• Force by which Body is subjected, F  = 20 N

To find :

• The kinetic energy acquired by the body at the end of 10 seconds ​.

Formulae required :

• Newton's second law of motion

$\red{\bigstar}\boxed{\sf{F=m\;a}}$

• First equation of motion

$\red{\bigstar}\boxed{\sf{v=u+a\;t}}$

• Formula to calculate Kinetic energy

$\red{\bigstar}\boxed{\sf{K.E.=\dfrac{1}{2}\;m\;v^2}}$

[ where F is force applied, m is mass, a is acceleration, v is final velocity, u is initial velocity, t is time taken by a body and K.E. is kinetic energy ]

Solution  :

Let, acceleration of the body = a

and, we need to find K.E. of body at , t = 10 s

so,

Using Newton's second Law of motion

$\longrightarrow\sf{F=m\;a}$

$\longrightarrow\sf{20=5\times a }$

$\longrightarrow\underline{\underline{\red{\sf{a=4\;ms^{-2}}}}}$

Now,

Using first equation of motion

$\longrightarrow\sf{v=u+a\;t}$

$\longrightarrow\sf{v=0+(4)\times (10)}$

$\longrightarrow\underline{\underline{\red{\sf{v=40\;ms^{-1}}}}}$

Using formula to Calculate K.E.

$\longrightarrow\sf{K.E.=\dfrac{1}{2}\;m\;v^2}$

$\longrightarrow\sf{K.E.=\dfrac{1}{2}\times(5)\times(40)^2}$

$\longrightarrow\sf{K.E.=\dfrac{1}{2}\times(5)\times(40)^2}$

$\longrightarrow\sf{K.E.=\dfrac{1}{2}\times 5\times 1600}$

$\longrightarrow\underbrace{\overbrace{\underline{\underline{\large{\red{\sf{K.E.=4000\;J}}}}}}}$

therefore,

• Kinetic Energy acquired by the body at the end of 10 seconds will be 4000 Joules .