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Question: A box of mass 2 kg is placed on the roof of a car. The box would remain stationary until the car att...

A box of mass 2 kg is placed on the roof of a car. The box would remain stationary until the car attains a maximum acceleration. Coefficient of static friction between the box and the roof of the car is 0.2 and g=.10m/s210m/{{s}^{2}} This maximum acceleration of the car, for the box to remain stationary is :
A. 8m/s28m/{{s}^{2}}
B. 6m/s26m/{{s}^{2}}
C. 4m/s24m/{{s}^{2}}
D. 2m/s22m/{{s}^{2}}

Explanation

Solution

When the car starts to move, the box on top experiences a static friction and the coefficient of static friction is nothing but the ratio between the maximum static force before motion to the normal force. We have to find the acceleration at which both these forces are equal.

Formulas used:
Equilibrium equation for the block: ma=μmgma=\mu mg
Where mmis the mass of the block, aa is the acceleration experienced , μ\mu is the coefficient of static friction and ggis the acceleration due to gravity.

Complete step by step answer:
We are given that the mass of the block m=2kgm=2kg and μ=0.2\mu =0.2
Also we know that g=10m/s2g=10m/{{s}^{2}}
In the given conditions, the equilibrium equation of the block is given by
ma=μmgma=\mu mg
Upon cancelling mm, we get
a=μga=\mu g
Upon substituting the given values, we obtain
a=0.2×10=2m/s2a=0.2\times 10=2m/{{s}^{2}}
Hence the maximum acceleration of the car for the box to remain stationary is 2m/s22m/{{s}^{2}}
Therefore, option D is the correct answer among the four.

Note: The value of μ\mu , i.e. the coefficient of static friction, depends on the objects that are causing friction. Since it’s the ratio, it is a dimensionless unit. Its value generally lies between 0 and 1 but sometimes it is greater than 1. When it is 1, the frictional force equals the normal force and when it is 0, there is no frictional force.