Question

A force acts on a $1\text{kg}$ particle such that position of particle as a function of time is given by $x=2t^2-4$, where ${}^{\prime }{x}^{\prime }$ in $\text{m}$ and $\text{t}$ is in $\text{seconds}$. The work done during initial $\text{4}$ seconds is

• A. $256\text{J}$
• B. $54\text{J}$
• C. $128\text{J}$
• D. $0\text{J}$

Answer 1

The correct option is C $128\text{J}$

Displacement of the particle is given by $x=2t^2-4$
Velocity obtained by the differentiation of the displacement w.r.t. time as
$v=\frac{dx}{dt}=\frac{d}{dt}(2t^2-4)=4t$

Velocity at time $t=0$ and Velocity at time $t=4\text{sec}$,
${\begin{array}{c}{v}_0\end{array}|}_{\text{at}t=0}=0\text{m/s}$
${\begin{array}{c}{v}_4\end{array}|}_{\text{at}t=4}=4\times 4=16\text{m/s}$

According to work energy theorem
Net work done = Change in KE
$=\frac{1}{2}\times 1(v_4^2-v_0^2)$
$=\frac{1}{2}\times 1({16}^2-0)$
$=128\text{J}$

Hence option C is the correct answer.


OR

Displacement of the particle is given by $x=2t^2-4$
Velocity obtained by the differentiation of the displacement w.r.t. time as
$v=\frac{dx}{dt}=\frac{d}{dt}(2t^2-4)=4t$
Acceleration $a=\frac{dv}{dt}=4$
Force acted upon the body is $F=ma=4\text{N}$

Displacement in initial $4$ seconds is $x_4-x_0=2(4{)}^2-4-(0-4)=32m$

Work done by constant force is $W=Fs=4\left(32\right)=128\text{J}$