Chapter 6, Problem 60A

Austin’s truck has a mass of 2000 kg. When traveling at 22.0 m/s, it brakes to a stop in 4.0 s. Show that the magnitude of the braking force acting on the truck is 11,000 N.

To determine

To show: Magnitude of braking force acting on truck is 11000 N.

Explanation of Solution

Given:

Mass of truck, $m=2000\text{kg}$

Time, $t=4\text{s}$

Initial velocity, $u=22\text{m/s}$

Final velocity, $v=0\text{m/s}$

Formula Used:

From the equation of motion

  $v=u+at$

Where, $v$ is initial velocity, $v$ is final velocity and $a$ is acceleration.

And from Newton’s second law, the net force is given as

  $F=ma$

Where, $m$ is mass of object and $a$ is acceleration.

Calculation:

On substituting the given values,

  $\begin{array}{l}0=22+a\times 4\\ \Rightarrow a=-\frac{22}{4}=-5.4{\text{m/s}}^2\end{array}$

As acceleration is negative, it means it is deceleration which is 5.4 m/s²

So, magnitude of acceleration is 5.4 m/s²

Now, the magnitude of breaking force is

  $\begin{array}{l}F=ma\\ =2000\times 5.5\\ =11000\text{N}\end{array}$

Conclusion:

Therefore, it is proved that the magnitude of braking force acting on the truck is 11000 N.