Question

A stone of mass $500g$ is thrown vertically upwards with a velocity of $15m{s}^{-1}$. Calculate

(i) The $PE$ at the greatest height.

(ii) $KE$ on reaching the ground.

(iii) The total energy at its half-way point.

Answer 1

Given data:

Mass $m=500g$

We know that $1kg=1000g$,therefore

$$\begin{gathered} m=500g \\ \Rightarrow \frac{500}{1000}kg \end{gathered}$$

$=0.500kg$

Velocity $v=15m{s}^{-1}$

(i) Calculation of $PE$ at the greatest height:

The stone is thrown vertically upwards ,therefore the potential energy at the greatest height is equal to the initial kinetic energy i.e.

Potential energy at the greatest height $=$ Initial kinetic energy $\left(1\right)$

$mgh=\frac{1}{2}m{v}^2$, where $m$ is mass, $g$ is acceleration due to gravity, $h$ is height and $v$ is velocity.

As to calculate using the formula $PE=mgh$,we don’t have $h$ (height) ,so we can calculate using the formula $KE=\frac{1}{2}m{v}^2$ as both are equal form $\left(1\right)$

$$\begin{gathered} PE=KE \\ \Rightarrow \frac{1}{2}m{v}^2 \end{gathered}$$

$=\frac{1}{2}\times 0.500\times {\left(15\right)}^2$

$PE=56.25J$ $\left(2\right)$

(ii) Calculation of kinetic energy on reaching the ground:

Kinetic energy ,is given by

$KE=\frac{1}{2}m{v}^2$

The value is already calculated ,that is from $\left(1\right)$ and $\left(2\right)$,we can conclude that,

$KE=56.25J$ $\left(3\right)$

Kinetic energy on reaching the ground is $56.25J$

(iii) Calculation of total energy at its half way:

Total energy at is half-way point$=\frac{1}{2}(KE+PE)$

Substitute the value form $\left(2\right)$ and $\left(3\right)$,we get

Total energy at is half-way point$=\frac{1}{2}(56.25+56.25)$

$=\frac{1}{2}(112.5)$

$=56.25J$

Hence, for a stone of mass $500g$ is thrown vertically upwards with a velocity of $15m{s}^{-1}$,

(i) Potential energy at the greatest height is $56.25J$.

(ii) Kinetic energy on the ground is $56.25J$.

(iii) Total energy at its half-way is $56.25J$.