Question

A $4kg$ particle moves along the $X-$axis. Its position $x$ varies with time according to $x\left(t\right)=t+2t^3$ , where $X$ is in $m$ and $t$ is in seconds. Compute kinetic energy of the particle in time t

• A. $KE=72t^2+24t+2$
• B. 0
• C. 20t
• D. 40t

Answer 1

The correct option is A $KE=72t^2+24t+2$

Kinetic energy is given by,

$KE=\frac{1}{2}m{v}^2$

$v=\frac{dx}{dx}$

$=\frac{d}{dt}(t+2t^3)$

$v=1+6t^2$

$\Rightarrow v^2=(1+6t^2{)}^2=1+36t^4+12t^2$

$KE=\frac{1}{2}\times 4\times (1+36t^4+12t^2)$

$=2(1+36t^4+12t^2)$

$\therefore KE=72t^2+24t+2$