Question

A force acts on a $10\text{g}$ particle in such a way that the position of particle as a function of time is given by $x=3+4t^2$, where $x$ is in $\text{m}$ and $\text{t}$ is in seconds. The work done during first $7$ seconds is

• A. $156.8\text{J}$
• B. $15.68\text{J}$
• C. $78.4\text{J}$
• D. $7.84\text{J}$

Answer 1

The correct option is B $15.68\text{J}$

Displacement is given by $x=3+4t^2$

We know that velocity is given by the time rate of change of displacement.

differentiation of $x$ w.r.t. time is $v=\frac{dx}{dt}=8t$

$v_0=8\times 0=0\text{m/s}$ [Velocity at $t=0Sec$]

$v_7=8\times 7=56\text{m/s}$ [Velocity at $t=7Sec$]

[Change in kinetic energy $=\frac{1}{2}\times m\times (v_7^2-v_0^2)$ ]
According to work energy theorem
Net work done $=$ Change in kinetic energy

$W=\frac{1}{2}\times \frac{10}{1000}\times \left({56}^2\right)$
$=15.68\text{J}$

Hence option B is the correct answer