# The coefficient of friction between the surface of a rock and ice is 0.1. A rock is slid at 40 m/s. If it comes to stop after 12 seconds, what is the mass of the rock.

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### 1 Answer

A rock is placed on a surface of ice. The coefficient of friction between the rock and the sheet of ice is 0.1. If the coefficient of friction between two surfaces is `mu` , the actual resistive force of friction between the two is equal to F = `N*mu` where N is the normal force. Here, the normal force of the rock and the ice surface is equal to M*9.8*0.1 where M is the mass of the rock.

The initial speed of the rock when it starts to slide on the ice is 40 m/s. It comes to a stop after 12 seconds. The deceleration of the rock is `(40 - 0)/12 = 10/3` m/s^2. The force resulting in the deceleration is `(10/3)*M` . This is equal to the frictional force. Equating the two expressions gives:

`(10/3)*M = M*9.8*0.1`

In the equation derived above M cancels on both the sides. In addition, `10/3 != 0.98`

The information provided in the problem is incorrect. The deceleration of the rock is not dependent on its mass and the value of the same is not what it should be from the coefficient of friction that is given.

*(Please correct the values provided)*