Question

A truck is carrying a box of mass \[m = 50\,kg\] on its flat horizontal rough surface with coefficient of friction \[p = 0.3\]. It is crossing a circular track of radius 28 m. What is the maximum speed of the truck to the box that does not slide from the muck while moving on the circular path?
A. \[18\,km/hr\]
B. \[36\,km/hr\]
C. \[32.4\,km/hr\]
D. \[2.5\,km/hr\]

Answer 1

Hint: The necessary centripetal force required to remain in the circular track for the box is provided by the friction force between the box and the truck. Equate these two forces and write the expression or the maximum velocity.

Complete step by step answer:
For the circular motion of the body, the necessary centripetal force is,
\[{F_c} = \dfrac{{m{v^2}}}{r}\]
Here, m is the mass of the body, v is the velocity and r is the radius of the circular motion.
On the circular track, the frictional force between the flat surface of the truck and the lower surface of the box balances the centripetal force on the box.
Therefore, we can write,
\[\dfrac{{m{v^2}}}{r} = pmg\]
\[\dfrac{{{v^2}}}{r} = pg\]
Here, p is the coefficient of the friction and g is the acceleration due to gravity.
Rearrange the above equation for \[v\].
\[v = \sqrt {prg} \]
This is the maximum velocity of the truck beyond which the box will skid off the truck.
Substitute 0.3 for p, 27 m for r and \[10\,m/{s^2}\] for g in the above equation.
\[{v_{\max }} = \sqrt {\left( {0.3} \right)\left( {27\,m} \right)\left( {10\,m/{s^2}} \right)} \]
\[ \Rightarrow {v_{\max }} = 9\,m/s\]
Convert the velocity into km/hr as follows,
\[{v_{\max }} = \left( {9\,m/s} \right)\left( {\dfrac{{1\,km}}{{1000\,m}}} \right)\left( {\dfrac{{3600\,s}}{{1\,hr}}} \right)\]
\[{v_{\max }} = 32.4\,km/hr\]

So, the correct answer is “Option C”.

Note:
To convert the velocity of the object from m/s into km/hr, multiply the velocity by \[3.6\,km/hr\].
Also remember that we have to apply the concept of centripetal force to remain in a circular path.