Question

A man drops a 10kg rock from the top of a 5m ladder, what is its speed just before it hits the ground? What is its kinetic energy when it reaches the ground? $\left( {g = 10m/{s^2}} \right)$

Answer 1

Hint: At the time when the man drops a 10kg rock from the top of a 5m ladder, the potential energy is converting into kinetic energy or we can look at it like this that after the rock touches the ground the kinetic energy is completely getting transferred into potential energy.

Complete step by step solution:
$K.E = mgh$ ;
Here,
m = Mass;
g = Gravitational Acceleration;
h = Height;
Now, we have been given the mass and the height,
$K.E = 10 \times 5 \times 10$;
$ \Rightarrow K.E = 500J$;
Now, the formula for kinetic energy is:
$K.E = \dfrac{1}{2}m{v^2}$ ;
Here: m = Mass;
v = velocity;
$500 = \dfrac{1}{2} \times 10 \times {v^2}$;
$ \Rightarrow 100 = {v^2}$;
Take the square root on the LHS:
$ \Rightarrow \sqrt {100} = v$;
$ \Rightarrow v = 10m/s$;

Therefore, its speed just before it hits the ground is 10 m/s and the kinetic energy is 500J.

Note: Here, we need to visualize the given situation before jumping on to the answer. Imagine that a 10kg is being carried by you on a ladder at a 5m height and then you dropped the stone, at the time you dropped the stone the potential energy is converted into kinetic energy and after the stone hits the ground the kinetic energy is converted into potential energy in either case the energy is being transferred. To find out the speed apply the formula for kinetic energy and solve for velocity v.