# A car has a mass of 1520 kg. While traveling at 30m/s, the driver applies the brakes to stop the...

## Question:

A car has a mass of 1520 kg. While traveling at 30m/s, the driver applies the brakes to stop the car on a wet surface with a 0.40 coefficient of friction. How far does the car travel before stopping?

## Conservation of Energy:

An isolated system has a constant total energy. Energy is neither created nor destroyed, it is only transferred from one form to another. When a moving object experiences friction, it slows down as it loses energy as heat due to the rubbing effect with the surface. This heat is due to the work done by the friction and it is equal the kinetic energy lost by the object.

When the driver applies brakes, there is friction between the car tires and the wet surface. The work done by the friction is equal to the amount of kinetic energy lost by the car. In our case, the car comes to a complete halt, therefore, it loses all its initial kinetic energy. Therefore, the work done by the friction is

\begin{align} Work &= K.E \\[0.3cm] &= \frac{1}{2}(1520 \ \rm kg) \left(30 \ \frac{\rm m}{\rm s} \right)^2 \left(\frac{\rm J}{\rm kg \cdot \rm m^2/s^2} \right) \\[0.3cm] &= 6.84 \times 10^5 \ \rm J \end{align}\\

Note that the work done by a constant force equals the product of the magnitude of the force and the distance moved by the object. In our case, the magnitude of the force of friction is equal to the product of the coefficient of friction and the normal force

\begin{align} f &= 0.40N \\[0.3cm] &= 0.40mg \\[0.3cm] &= 0.40(1520 \ \rm kg) \left(9.81 \ \frac{\rm m}{\rm s^2} \right) \\[0.3cm] &= \ 5.96448 \times \ 10^3 \ \rm N \end{align}\\

Therefore, the work done by the friction is

\begin{align} Work &= fd \\[0.3cm] 6.84 \times \ 10^5 \ \rm N \cdot \rm m &= \left(5.96448 \times \ 10^3 \ \rm N \right) d \\[0.3cm] \end{align}

Solving for {eq}d{/eq}, the distance moved by the car before stopping is

\begin{align} d &= \frac{6.84 \times 10^5 \ \rm N \cdot m }{5.96448 \times 10^3 \ \rm N} \\[0.3cm] &\approx \boxed{115 \ \rm m} \end{align}