Question

A car weighing 1000kg and travelling at 30m/s stops at a distance of 50m decelerating uniformly. What is the force exerted by the brakes and what is the work done by the brakes?

Answer 1

Question#19135

A car weighing $1000\,\mathrm{kg}$ and travelling at $30\,\mathrm{m/s}$ stops at a distance of $50\,\mathrm{m}$ decelerating uniformly. What is the force exerted by the brakes and what is the work done by the brakes?

Solution:

Let:

$$m = 1000\,\mathrm{kg}$$ $$v = 30\,\mathrm{m/s}$$ $$S = 50\,\mathrm{m}$$

F—?

According to the second Newton’s law:

$F = ma$, were a–acceleration

$$v = at, t = \frac{v}{a}, S = \frac{1}{2}at^2$$ $$S = \frac{1}{2}\frac{v^2}{a}$$ $$a = \frac{v^2}{2S}$$ $$F = \frac{mv^2}{2S}$$ $$F = \frac{1000 \cdot 30^2}{2 \cdot 50} = 9000\,\mathrm{N}$$

Answer: 9000 N.