Question

A man drops a 10 kg rock from the top of a 5 m ladder. What is its speed just before it hits the ground?
What is its kinetic energy when it reaches the ground ? (g = 10 m/s)​

Answer 1

Given :

• A man drops a 10 kg rock from the top of a 5 m ladder.

To Find :

Kinetic energy of rock when it reaches the ground

Theory :

• Work energy theorem

The change in kinetic energy of a particle is equal to the work done by the net force.

$\bf\green{W=\triangle\:K.E.}$

• Law of conservation of mechanical energy

The sum of the potential energy and kinetic energy is called the total mechanical energy.

$\bf\blue{\triangle\:U+\triangle\:K.E.}$

Solution:

We have to find Kinetic energy of rock when it reaches the ground.

By Work energy theorem

$\sf{W=\triangle\:K.E.}$

$\sf\implies\:W_{Gravity}=K.E_{final}-KE_{initial}$

$\sf\implies\:mgh=\dfrac{1}{2}mv^2-0$

Put the given values

$\sf\implies\dfrac{1}{2}mv^2=10\times10\times5$

$\sf\implies\:KE_{final}=500J$

Therefore , Kinetic energy of rock when it reaches the ground is 500 J.

Answer 2

Given:-

• A man drops a 10 kg rock from the top of a 5 m ladder.

To Find:-

• Kinetic energy of rock when it reaches ground.

Solution:-

By work energy theorem

$\sf{W\:=\:∆K.E.}$

$\sf\implies{W_Gravity\:=\:K.E_final-K.E_initial}$

$\sf \implies \: mgh \: = \: \frac{1}{2} {mv}^{2} - 0$

Put the given values

$\sf \implies\frac{1}{2} {mv}^{2} \: = \: 10 \times 10 \times 5$

$\sf\implies{KE_final\:=\:500J}$

Hence, Kinetic Energy of rock when it reaches the ground is 500J.

Additional Information:-

1) work and energy are interchangeable quantities

2) when the speed of the particle is constant there is no change in kinetic energy.