Question

A force given by the relation $\mathrm{F}=8\mathrm{t}$ acts on a body of mass $2\mathrm{kg}$, initially at rest. Find the work done by this force on the body during the first $2$ seconds of its motion.

• A. $0$
• B. $1$
• C. $-64\mathrm{J}$
• D. $64\mathrm{J}$

Answer 1

The correct option is D

$64\mathrm{J}$

Step 1: Given data

The force is $\mathrm{F}=8\mathrm{t}$,

Mass $\mathrm{m}=2\mathrm{kg}$

We have to find work done.

Step 2: Formula to be used

Newton's second law states that force $\left(\mathrm{F}\right)$ acting on an object is equal to the mass $\left(\mathrm{m}\right)$ times acceleration$\left(\mathrm{a}\right)$.

So,

$\mathrm{F}=\mathrm{m}\times \mathrm{a}$

Step 3: Find the work done

As we know, work done $=$ Force $\times$ Distance $\cos \mathrm{\theta }$

Here, $\mathrm{\theta }=0°$

So,

$\cos \left(0°\right)=1$

Step 4: Find the distance travelled

We know that,

$\mathrm{F}=\mathrm{m}\times \mathrm{a}$

So,

$\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}$

The distance travelled in $2$ seconds is,

Acceleration $=\frac{8\mathrm{t}}{2\mathrm{kg}}$

$=4\mathrm{m}/{\mathrm{s}}^2$

According to the equation of motion,

$\mathrm{S}=\mathrm{ut}+0.5\times {\mathrm{at}}^2$

$=0+0.5\times 4\times 2\times 2$

$=8\mathrm{m}$

Step 4: Find work done by force

Work done by the force is,

$\mathrm{FS}=8\times 8$

$=64\mathrm{J}$

So, the work done is $64\mathrm{J}$.

Hence, option D is correct.