Question

A Certain force acting on a 20 kg man changes its Velocity from 5m/s to 2m/s. Calculate the work done by the force.

Please help me solve this sum. I will mark you brainliest. ​

Answer 1

Mass of the object=20kg

initial velocity=5m/s

final velocity=2m/s

Work done=change in kinetic energy

Initial k.n=1/2mu²

=1/2x20x5²

=10x25

=250J

Final k.n=1/2mu²

=1/2x20x2²

=10x4

=40J

Work done=Final kinetic energy-initial kinetic energy

=40J-250J

=-210J

Answer 2

Answer:

• The work done (W) is - 210 J

Given:

1. Mass of man (M) = 20 Kg
2. Initial velocity (u) = 5 m/s
3. Final velocity (v) = 2 m/s.

Explanation:

$\rule{300}{1.5}$

From work - energy theorem we know,

$\bigstar\;\underline{\boxed{\sf W=\dfrac{1}{2}\;M\;\bigg(v^{2} - u^{2}\bigg)}}$

Here,

• W Denotes work.
• v Denotes final velocity.
• u Denotes initial velocity.

Substituting the values,

$\longrightarrow\sf W=\dfrac{1}{2}\times 20\times \Bigg[\bigg(2\bigg)^{2}-\bigg(5\bigg)^{2}\Bigg]\\\\\\\longrightarrow\sf W=\dfrac{1}{2}\times 20\times \Bigg[4-25\Bigg]\\\\\\\longrightarrow\sf W=\dfrac{1}{2}\times 20\times -21\\\\\\\longrightarrow\sf W=10\times - 21\\\\\\\longrightarrow\sf W=-210\\\\\\\longrightarrow \large{\underline{\boxed{\red{\sf W=-210\;J}}}}$

∴ The work done (W) by the force is - 210 J

Note:-

• The negative sign indicates that Force is opposing the motion of the man.

$\rule{300}{1.5}$