Problem

# Energy is a conserved quantity and Quantifying gravitational potential and kinetic e...

Energy is a conserved quantity and Quantifying gravitational potential and kinetic energies

A pickup truck (2268 kg) and a compact car (1100 kg) have the same momentum. (a) What is the ratio of their kinetic energies? (b) If the same horizontal net force were exerted on both vehicles, pushing them from rest over the same distance, what is the ratio of their final kinetic energies?

#### Step-by-Step Solution

Solution 1

The gravitational potential energy is nothing but the energy of the object-earth system associated with the elevation of the object or a body above the Earth.

The kinetic energy is nothing but the energy due to the motion of the object or a body.

The elastic potential energy is nothing but the energy associated with an elastic object's degree of stretch.

The expression for the kinetic energy of an object is,

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

Here, $$m$$ is the mass of the object and $$v$$ is the speed of the object.

(a)

The magnitude of momentum of an object of mass $$m$$ moving at speed $$v$$ is,

$$p=m v$$

Rewrite the equation for $$v$$.

$$v=\frac{p}{m}$$

The kinetic energy of the object is,

\begin{aligned} K &=\frac{1}{2} m\left(\frac{p}{m}\right)^{2} \\ &=\frac{p^{2}}{2 m} \end{aligned}

The kinetic energy of the pickup truck is,

$$K_{p}=\frac{p^{2}}{2 m_{p}}$$

The kinetic energy of the compact car is,

$$K_{c}=\frac{p^{2}}{2 m_{c}}$$

The ratio of kinetic energies is,

\begin{aligned} \frac{K_{\mathrm{p}}}{K_{\mathrm{c}}} &=\frac{p^{2} / 2 m_{\mathrm{p}}}{p^{2} / 2 m_{\mathrm{c}}} \\ &=\frac{m_{\mathrm{c}}}{m_{\mathrm{p}}} \end{aligned}

Substitute $$1100 \mathrm{~kg}$$ for $$m_{c}$$ and $$2268 \mathrm{~kg}$$ for $$m_{p}$$.

\begin{aligned} \frac{K_{\mathrm{p}}}{K_{\mathrm{c}}} &=\frac{1100 \mathrm{~kg}}{2268 \mathrm{~kg}} \\ &=0.485 \end{aligned}

Therefore, the ratio of kinetic energies is $$0.485$$.

Comment

Step 3 of 3 人

(b)

The work done is equal to the change in kinetic energy. That is,

$$\begin{array}{r} W=F d \\ \Delta K=F d \end{array}$$

Here, the same magnitude of force and same distance displaced, we have $$K_{\mathrm{p}}=K_{\mathrm{c}}$$. Hence, the ratio of kinetic energies is 1 .