Question

A 1400 kg car is traveling at 24 m/s when the driver takes his foot off of the gas. The car eventually rolls to a stop after 225 m. Find the force of friction acting on the car.

Answer 1

Answer on Question #62845, Physics / Mechanics | Relativity

Question:

A 1400 kg car is traveling at 24 m/s when the driver takes his foot off of the gas. The car eventually rolls to a stop after 225 m. Find the force of friction acting on the car.

Solution:

Let's suppose that kinetic energy of the car is consumed entirely by friction. According to the law of conservation of energy we may state that the work of the force of friction is equal to initial kinetic energy of the car.

$$= \frac{m v^{2}}{2}, \text{ where } \quad - \text{kinetic energy;}$$ $$m - \text{the mass of the car;}$$ $$v - \text{initial speed of the car.}$$ $$W = F \cdot s, \text{ where } W - \text{the work of the force of friction;}$$ $$F - \text{the force of friction;}$$ $$s - \text{displacement of the car.}$$

In our case $= W$, that is $\frac{m v^{2}}{2} = F \cdot s$ and we may express $F = \frac{m v^{2}}{2s}$.

$$F = \frac{1400 \cdot 24^{2}}{2 \cdot 225} = 1792 \, \text{N}$$

Answer:

1792 N

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