A cup of coffee is sitting on a table in a train that is moving with a constant velocity. The coefficient of static friction between the cup and the table is 0.30. Suddenly, the train accelerates. What is the maximum acceleration that the train can have without the cup sliding backward on the table?

a = 2.94 m/s²

Explanation:

In order for the cup not to slip, the unbalanced force on cup must be equal to the frictional force:

Unbalanced Force = Frictional Force

ma = μR = μW

ma = μmg

a = μg

where,

a = maximum acceleration for the cup not to slip = ?

μ = coefficient of static friction = 0.3

g = acceleration due to gravity = 9.8 m/s²

Therefore,

a = (0.3)(9.8 m/s²)

a = 2.94 m/s²

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