Question

A block of mass $5\text{kg}$ is moving in $x-$ direction with a constant speed of $20\text{m/s}$. It is subjected to a retarding force of $F=-0.1x\text{joules/metre}$ during its travel from $x=20\text{m}$ to $x=30\text{m}$. Its final K.E. will be:

• A. $975\text{J}$
• B. $1025\text{J}$
• C. $825\text{J}$
• D. $725\text{J}$

Answer 1

The correct option is A $975\text{J}$

Since the block is moving under the action of a variable force, work done by the force for displacement from $x=20\text{m}$ to $x=30\text{m}$ is given by
$W=\underset{x_i}{\overset{x_f}{\int }}Fdx=\underset{20}{\overset{30}{\int }}(-0.1x)dx$
$W=-0.1{\left[\frac{x^2}{2}\right]}_{20}^{30}=-\frac{0.1}{2}\left[\right(30{)}^2-\left(20{)}^2\right]$
$\therefore W=-25\text{joules}$

Now, initial kinetic energy $K{E}_i=\frac{1}{2}m{u}^2$
$K{E}_i=\frac{1}{2}\times 5\times (20{)}^2=1000\text{J}$
Let $K{E}_f$ be the final kinetic energy.
Applying work energy theorem on the block:
$W_{\text{ext}}=\Delta KE$
$\Rightarrow W=K{E}_f-K{E}_i$
$\Rightarrow -25=K{E}_f-1000$
$\therefore K{E}_f=1000-25=975\text{J}$