If an object is moving at #15 m/s# over a surface with a kinetic friction coefficient of #u_k=225 /g#, how much time will it take for the object to stop moving?

Question

If an object is moving at #15 m/s# over a surface with a kinetic friction coefficient of #u_k=225    /g#, how much time will it take for the object to stop moving?

Isabella Alvarez · Accepted Answer

#=1/15s# we know that frictional force acting on the body while moving in horizontal surface is given by Kinetic friction #F_k=mu_kmg#,where m= mass and g = acceleration due to gravity So the retardation a#=F_k/m =(mu_kmg)/m=mu_kg=225/g*g=225 ms^-2# Initial velocity of the body #u =15ms^-1# Final velocity v = 0 If time required to stop be t then #v=u-at# #=>0=15-225t# #=>t=15/225s=1/15s#

William Audy · Answer

undefined To calculate the time it takes for the object to stop moving, you can use the equation $ t = \frac{v}{\mu_k g} $, where $ t $ is the time, $ v $ is the initial velocity, $ \mu_k $ is the kinetic friction coefficient, and $ g $ is the acceleration due to gravity.

Given:
- Initial velocity, $ v = 15 \, \text{m/s} $
- Kinetic friction coefficient, $ \mu_k = 225 \, \text{/g} $ (units must be clarified)

The acceleration due to gravity is approximately $ 9.8 \, \text{m/s}^2 $.

Plugging in the values:

$ t = \frac{15}{225 \times 9.8} $

$ t \approx 0.0065 $ seconds.