Question

Under the action of a force, a $2$ kg body moves such that its position ${}^{\prime }{x}^{\prime }$ in meters as a function of time ${}^{\prime }{t}^{\prime }$ in seconds given by: $x=t^2/2$. The work done by the force in the first $5$ seconds is?

Answer 1

Mass= $2$ kg

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

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

Acceleration = $\frac{dv}{dt}=\frac{d^{2}x}{{dt}^2}$ = $1\frac{m}{s^2}$

$F=\left(m\right)\left(a\right)=\left(2\right)\left(1\right)=2N$

Work done in 5 sec,

$W=\left(F\right)\left(s\right)$ = $2\times \frac{t^2}{2}=2\times \frac{5^2}{2}$ J = $25$ J