Question

The kinetic energy of an object of mass, m moving with a velocity of $5{\mathrm{ms}}^{-1}$ is $25\mathrm{J}$, What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times?

Answer 1

Step 1: Given data

The kinetic energy of an object (K.E) = $25\mathrm{J}$

The velocity of the object (V) = $5{\mathrm{ms}}^{-1}$

Mass of the object = m

Step 2: Formula used

$\mathrm{K}.\mathrm{E}=\frac{1}{2}{\mathrm{mv}}^2$

For the mass of the object

$\begin{array}{rcl}& \Rightarrow & \mathrm{K}.\mathrm{E}=\frac{1}{2}{\mathrm{mv}}^2\\ & \Rightarrow & 25=\frac{1}{2}\times \mathrm{m}\times 5^2\\ & \Rightarrow & 50=25\mathrm{m}\\ & \Rightarrow & \mathrm{m}=2\mathrm{kg}\end{array}$

Step 3: Calculation

Case 1:

When velocity is doubled, then

$\begin{array}{rcl}\mathrm{v}& =& 10{\mathrm{ms}}^{-1}\\ \mathrm{m}& =& 2\mathrm{kg}\end{array}$

So

$\begin{array}{rcl}& \Rightarrow & \mathrm{K}.\mathrm{E}=\frac{1}{2}{\mathrm{mv}}^2\\ & \Rightarrow & \mathrm{K}.\mathrm{E}=\frac{1}{2}\times 2\times {10}^2\\ & \Rightarrow & \mathrm{K}.\mathrm{E}=100\mathrm{J}\end{array}$

So its kinetic energy is $100\mathrm{J}$ when its velocity is doubled.

Case 2:

When its velocity is increased three times, then

$\begin{array}{rcl}\mathrm{v}& =& 15{\mathrm{ms}}^{-1}\\ \mathrm{m}& =& 2\mathrm{kg}\end{array}$

So

$\begin{array}{rcl}& \Rightarrow & \mathrm{K}.\mathrm{E}=\frac{1}{2}{\mathrm{mv}}^2\\ & \Rightarrow & \mathrm{K}.\mathrm{E}=\frac{1}{2}\times 2\times {15}^2\\ & \Rightarrow & \mathrm{K}.\mathrm{E}=225\mathrm{J}\end{array}$

So its kinetic energy is $225\mathrm{J}$ when its velocity is increased three times.