Question

Coefficient of friction the flatbed of the truck and crate are ${{\mu }_{s}}=0.8$ and ${{\mu }_{k}}=0.7$. The coefficient of kinetic friction between the truck tyres and the road surface is 0.9. If the truck stops from an initial speed $15m{{s}^{-1}}$ with maximum braking (wheels skidding). Determine where on the bed the crate finally comes to rest. (Take $g=10m{{s}^{-2}}$).

A. 2.77m
B. 3.57m
C. 3m
D. 1.77m

Answer 1

Hint:Use the formula for frictional force and newton’s second law of motion and calculate the retardation of the truck and the crate. Later, by using kinematic equations, find the displacement of both. Then you can calculate the displacement of the crate with respect to the truck.

Formula used:
$f={{\mu }_{k}}N$
$2as={{v}^{2}}-{{u}^{2}}$
where a, s, v and u are the acceleration, displacement, final and initial velocities of the body.

Complete step by step answer:
It is given that the truck stops with maximum braking. This means that the tyres of the truck stop rolling at once and skid without rolling on the ground. Then the truck will come to rest due to the kinetic friction force between the tyres and the ground. Kinetic friction on a body given as $f={{\mu }_{k}}N$, where ${{\mu }_{k}}$ is the coefficient of kinetic friction between the body and the surface in contact. N is the normal reaction exerted on the body.

In this case, the normal reaction on the truck is $N=Mg$, where M is the sum of the masses of the truck and the crate on it. And it is given that the coefficient of kinetic friction between the truck tyres and the road surface is 0.9. Therefore, the frictional force on the truck is ${{f}_{T}}={{\mu }_{k}}Mg=(0.9)M(10)=9M$. From Newton’s second law of motion we know that ${{f}_{T}}=Ma$, where a is the acceleration of the truck. In this case, it is retardation.
$\Rightarrow {{f}_{T}}=Ma=9M$
$\Rightarrow a=-9m{{s}^{-2}}$ (since it is retarding)

Since the truck is retarding, there will be some relative motion between the crate and the truck, due to which the crate will also feel a frictional force. Now, we know that the truck is accelerating in the direction of the negative x-axis. Therefore, with respect to the truck, the crate will accelerate in the direction of positive x-axis. Thus, to oppose this motion, friction will exert a force on the crate in the direction of the negative x-axis. As a result, after some time the crate will also come to rest.

Let the mass of the crate be m. Therefore, the normal reaction of the carte is $N'=mg=10m$.
This means that frictional force on the crate is ${{f}_{C}}={{\mu }_{k}}N=(0.7)(10m)=7m$.
Let the acceleration of the crate be a’. Then we can write that ${{f}_{C}}=ma'$.
$\Rightarrow {{f}_{C}}=ma'=7m$
$\Rightarrow a'=-7m{{s}^{-2}}$ (since it is retarding)
This means that the crate is retarding at a rate of $7m{{s}^{-2}}$.
Now, we shall use the kinematic equation $2as={{v}^{2}}-{{u}^{2}}$ and find the displacement of the crate.
Here, $s={{s}_{c}},u=15m{{s}^{-1}},v=0,a=-7m{{s}^{-2}}$

After substituting these values,
$\Rightarrow 2(-7){{s}_{c}}={{0}^{2}}-{{15}^{2}}$
$\Rightarrow {{s}_{c}}=16.07m$.
Now using the same kinematic equation we shall find the displacement of the truck.
Substitute $s={{s}_{T}},u=15m{{s}^{-1}},v=0,a=-9m{{s}^{-2}}$.
$\Rightarrow 2(-9){{s}_{T}}={{0}^{2}}-{{15}^{2}}$
$\Rightarrow 2(-9){{s}_{T}}=12.5m$
Therefore, the displacement of the crate with respect to the truck is ${{s}_{c}}-{{s}_{T}}=16.07-12.5=3.57m$

Hence, the correct option is B.

Note:Please note that here we have neglected the frictional force that is exerted on the truck due to the friction between crate and the truck. This is because the mass of the truck is much larger than the mass of the crate. As a result, the friction between crate and truck is much less than the friction between truck and ground. Hence, can be neglected.