# A block weighing 180 kg is moving on a horizontal surface at 20 m/s initially. If the coefficient of kinetic friction is 1.8, what distance does the block move before it stops?

*print*Print*list*Cite

### 1 Answer

The mass of the block is 180 kg and the coefficient of kinetic friction between the block and the surface it is moving on is 1.8. The initial speed of the block is 20 m/s.

The force of friction is calculated as F = N*Cf, where N is the normal force and Cf is the coefficient of kinetic friction. Substituting the values we have F = 180*9.8*1.8.

Due to the resistive force of friction, there is a negative acceleration on the block that is equal to 180*9.8*1.8/180 = 9.8*1.8 = 17.64 m/s^2

For the block to come to a stop, the final velocity should be 0. Use the formula v^2 - u^2 = 2*a*s, where v and u are the final and initial velocities and s is the distance traveled.

0 - 20^2 = -2*17.64*s

=> s = 400/(2*17.64)

=> s = 11.33 m

The block moves for 11.33 m before it comes to a halt.