A system is devised to exert a constant force of 8 N on an 80-kg body of mass initially at rest. The force pushes the mass horizontally on a frictionless table. How far does the body have to be pushed to increase its mass-energy by 25%?
The net work done \((W)\) by the conservative force is equal to the change in kinetic energy \((\Delta K)\)
$$ W=\Delta K \ldots \ldots .(1) $$
If \(F\) is the applied force and \(d\) is the distance moved then the work done is \(W=F \cdot d \ldots \ldots\) (2)
Using equations (1) and (2),
$$ F \cdot d=\Delta K $$
Rearranging the terms in the above equation, the distance, \(d\) is
$$ d=\frac{\Delta K}{F} \ldots \ldots .(3) $$
The rest mass of the particle is \(m c^{2}\).
Here, mass of the particle is \(m\) and velocity of the light, \(c=3.0 \times 10^{8} \mathrm{~m} / \mathrm{s}\) Given that the increase in rest mass energy is
$$ \begin{aligned} \Delta K &=25 \%\left(m c^{2}\right) \\ &=\left(\frac{25}{100}\right)\left(m c^{2}\right) \\ &=0.25 m c^{2} \end{aligned} $$
Mass of the person is \(m=80 \mathrm{~kg}\)
The force exerted by the person is \(F=8.0 \mathrm{~N}\)
Using equation (3), the distance, \(d\) pushed is by the force is
$$ \begin{aligned} d &=\frac{\Delta K}{F} \\ &=\frac{0.25 m c^{2}}{F} \\ &=\frac{0.25(80.0 \mathrm{~kg})\left(3.0 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)^{2}}{8.0 \mathrm{~N}} \\ &=2.25 \times 10^{17} \mathrm{~m} \end{aligned} $$