Problem

# A system is devised to exert a constant force of 8 N on an 80-kg body of mass initially...

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%?

#### Step-by-Step Solution

Solution 1

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}