Question

A 8,000 kg amazon truck carrying a 3000 kg package is moving at 20 m/s along a level road as shown to the left. The driver applies the brakes to slow down. The truck travels 50 meters while applying the brakes for 3.0 seconds. The package does not slide on the back of the truck. a) Calculate the rate of acceleration b) Calculate the minimum coefficient of friction between packages and the truck

Answer 1

Answer:

(a) -6.67 $m/s^2$

(b) 0.68

Explanation:

Given that the mass of the truck, $m_1=8000$ kg.

Mass of the package, $m_2=3000$ kg.

As the package does not slide, so the acceleration of both, truck as well as the package, is the same.

Let $a\;\; m/s^2$ is the acceleration of the combined mass, m.

$m=m_1+m_2= 8000+3000=11000$ kg.

The initial velocity of the combined mass, u= 20 m/s.

Time required to stop, t=3 seconds.

Final velocity, v=0.

Displacement traveled, s=50 m.

(a) As  $a=\frac {v-u}{t}$

$\Rightarrow a=\frac{0-20}{3}=-\frac{20}{3} =6.67 m/s^2$

Hence the acceleration of the truck is $-6.67 m/s^2.$

(b) Now, $a=-6.67 m/s^2$ is the acceleration of the package, this acceleration is due the frictional force, f.

Due to inertia, on application of break, the package have tendency to slide to left (in  the direction of velocity). But the package does not slides, this is only due to the frictional force, f, which acts in the right direction ( opposite to the direction of velocity).

So, the magnitude of frictional force required on $m_2$  to avoid slide is

$f=|m_2a|=\frac{20}{3}m_2\times(i)$

Now, let \mu be the minimum coefficient of the friction, so

The force due to friction, $f_r= \mu N$,

where $\mu N$ is the normal reaction.

$N=m_2g$, where g is the acceleration due to gravity.

So, $f_r=\mu m_2g\cdots(ii)$

To avoid slide, $f_r\geq f$

$\Rightarrow \mu m_2g\geq\frac{20}{3}m_2$ [from (i)and (ii)]

$\Rightarrow \mu\geq \frac{20}{3\times 9.81}$    $[ as \;g=9.81 m/s^2]$

$\Rightarrow \mu \geq 0.68$

Hence, the minimum value of coefficient of friction between packages and the truck os 0.68.