A car with mass 1300 kg is traveling at 13.3 m/s. If the frictional force is 7700 N, what is the stopping distance?

$$W_{net} = \Delta KE$$ $$W_{net} = KE_f - KE_i$$

The formula for kinetic energy is $KE = \frac{1}{2}mv^2$.
Initial kinetic energy:

$$KE_i = \frac{1}{2}mv_i^2$$ $$KE_i = \frac{1}{2}(1300 \text{ kg})(13.3 \text{ m} \cdot \text{s}^{-1})^2$$ $$KE_i = \frac{1}{2}(1300 \text{ kg})(176.89 \text{ m}^2 \cdot \text{s}^{-2})$$ $$KE_i = 650 \text{ kg} \cdot 176.89 \text{ m}^2 \cdot \text{s}^{-2}$$ $$KE_i = 114978.5 \text{ J}$$

Final kinetic energy:

$$KE_f = \frac{1}{2}mv_f^2$$

Since the car stops, $v_f = 0 \text{ m} \cdot \text{s}^{-1}$.

$$KE_f = \frac{1}{2}m(0)^2 = 0 \text{ J}$$

The work done by friction is $W_f = F_f \cdot \Delta x \cdot \cos(\theta)$. Since the frictional force opposes the motion, the angle $\theta$ between the force and displacement $\Delta x$ is $180^\circ$, so $\cos(180^\circ) = -1$.

$$W_f = -F_f \cdot \Delta x$$ $$W_f = -(7700 \text{ N}) \cdot \Delta x$$

$$W_{net} = KE_f - KE_i$$ $$-F_f \cdot \Delta x = 0 - KE_i$$ $$-(7700 \text{ N}) \cdot \Delta x = -114978.5 \text{ J}$$ $$(7700 \text{ N}) \cdot \Delta x = 114978.5 \text{ J}$$ $$\Delta x = \frac{114978.5 \text{ J}}{7700 \text{ N}}$$ $$\Delta x = 14.9322727... \text{ m}$$

$$\Delta x \approx 14.93 \text{ m}$$

The stopping distance of the car is:

$$\boxed{14.93 \text{ m}}$$

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