Question

A truck of mass 1800 kg is moving with a speed of 54 km/h when brakes are applied it stops with uniform retardation at a distance of 200 m. Calculate the force applied by the brakes of the truck and the work done before stopping.

Answer 1

The initial velocity of the truck is,
$$\begin{gathered} u=54km/hr \\ u=54\times \frac{5}{18}m/s \\ u=15m/s \end{gathered}$$

Now, after applying brakes, the truck comes into rest position. thus, the final velocity of the truck will be $v=0$.
The acceleration of the truck will be found by using equation of motion as,
$v^2-u^2=2as$
On substituting the values we get,
$$\begin{gathered} {\left(0\right)}^2-{\left(15m/s\right)}^2=2a\times 200m \\ -{\left(15m/s\right)}^2=a\times 400m \\ a=-\frac{225m^2/s^2}{400m} \\ a=-0.56m/s2 \end{gathered}$$
Negative sign shows that the retardation takes place when brakes are applied.
Now, using Newton's second law of motion the retarding force will be calculated as,
$F=ma$
on substituting the values we get,
$$\begin{gathered} F=\left(1800kg\right)\times \left(-0.56m/s^2\right) \\ F=-1012.5N \end{gathered}$$
Thus, the magnitude of retarding force will be, $F=1012.5N$

Now, the work done is defined as the product of force and the displacement,
$W=Fs$
On substituting the values we get,
$$\begin{gathered} W=1012.5N\times 200m \\ W=202500N.m \end{gathered}$$