Question

The distance $\left(x\right)$ converted by a body of $2kg$ under the action of a force is related to time $t$ as $x=t^2/4$. What is the work done by the force in first $2$ seconds?

• A. $4J$
• B. $2J$
• C. $1J$
• D. $0.5J$

Answer 1

The correct option is C $1J$

Given that,

Mass $m=2kg$

Distance $x=\frac{t^2}{4}$

Now, on differentiate

$\frac{dx}{dt}=\frac{1}{2}t$

Now, velocity and acceleration is

$v=\frac{dx}{dt}=\frac{t}{2}$

$a=\frac{dv}{dt}=\frac{1}{2}$

Now, the force is

$F=ma$

$F=2\times \frac{1}{2}$

$F=1N$

Now, the work done is

$dW=F\cdot dx$

$\int dW=\underset{0}{\overset{2}{\int }}1\times \frac{t}{2}dt$

$W={\left[\frac{t^2}{4}\right]}_0^2$

$W=[\frac{4}{4}-0]$

$W=1J$

Hence, the work done is $1J$