If a #13 kg# object moving at #7 m/s# slows down to a halt after moving #10 m#, what is the friction coefficient of the surface that the object was moving over?

Question

Layla Ahmed · Accepted Answer

#mu_k=0.25#

I'll walk you through two different ways to answer this question:

Method 1: Applying Equations of Motion and Newton's Laws:

Using a pertinent equation of motion for constant acceleration in one dimension, the uniform and one-directional acceleration of the object can be determined as follows:

#v^2=u^2+2ax#

#therefore a=(v^2-u^2)/(2x)=(0^2-7^2)/(2xx10)=-2.45m//s^2#.

The frictional force, which is the resultant force acting on the object, is what causes this acceleration. Using Newton 2 and the definition of frictional forces, we obtain:

#sumvecF=mveca#

#therefore -f_k=ma#, where #f_k=mu_kN#.

But #N=mg =>mu_kmg=ma#.

#therefore mu_k=a/g=2.45/9.8=0.25#.

Method 2: Applying energy-related factors

According to the Work-Energy Theorem, the work that the resulting force accomplishes is equal to the kinetic energy change that occurs.

W_(f_k)=DeltaE_k, hence

#therefore f_k*x=1/2m(v^2-u^2)#

Therefore, 1/2m(v^2-u^2)=mu_kmg*x.

#therefore mu_k=(v^2-u^2)/(2gx)=(0^2-7^2)/(2xx9.8xx10)=0.25#.

Charles Aeschliman · Answer

undefined To find the friction coefficient, you can use the equation: friction force = coefficient of friction * normal force. The friction force can be calculated using the object's initial and final velocities, mass, and distance traveled. Then, use Newton's second law to find the normal force. Finally, divide the friction force by the normal force to find the coefficient of friction.