Question

A crate is on the flat surface of a truck of coefficient of static friction ${\mu }_s=0.8$ and coefficient of kinetic friction ${\mu }_k=0.7$. The coefficient of kinetic friction between the truck tires and the road surface is $0.9$. If the truck suddenly stops from an initial speed of $15m/s$ with maximum braking (wheels skidding), determine the displacement of the crate before it comes to rest.

• A. $3.2m$
• B. $2m$
• C. $2.79m$
• D. $1m$

Answer 1

The correct option is C $2.79m$

The FBD of the crate after the brakes are applied is given by


From the FBD, we have
$N=mg$
$f_{k1}={\mu }_{k1}N=0.7mg$

Retadation of crate is
$a_1=\frac{f_{k1}}{m}=\left(0.7\right)g=7m/s^2$

Assuming that mass of truck >> mass of crate, retardation of truck is only due to friction with the road.
Retardation of truck is
$a_2={\mu }_{k2}g=\left(0.9\right)g=9m/s^2$

Relative acceleration of crate w.r.t truck is
$a_r=(a_2-a_1)=2m/s^2$

Now, crate will strike the wall after time
$t_1=\sqrt{\frac{2s}{a_r}}=\sqrt{\frac{2\times 3.2}{2}}=1.79s$
and, truck will stop after time
$t_2=\frac{u}{a_2}=\frac{15}{9}=1.67s$

Since, $t_1>t_2$ , crate will come to rest before striking the front wall of the truck after travelling a distance
$s=\frac{1}{2}{a}_r{t}_2^2=\frac{1}{2}\times 2\times {\left(1.67\right)}^2=2.79m$