If a #8 kg# object moving at #16 m/s# slows to a halt after moving #80 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

2 Answers
Jan 20, 2018

Answer:

0.16

Explanation:

Clearly,here the frictional force caused retardation of the moving object and brought it to rest.

So,using #v^2 = u^2 - 2as # (all symbols are bearing their conventional meaning)(given, #v=0# ,#u=16 m/s# and #s = 80#)

We get, #a = 1.6 m/s^2#

Now,if the coefficient of kinetic friction is #mu# , frictional force(#f#) acting on the body was #mu*N# or #mu*mg# i.e #mu*80#

So,we using # f = ma # (where, #m=8Kg # and #a = 1.6 m/s^2#)

We, get #mu*80 = 12.8#
Or, # mu = 0.16#

Jan 20, 2018

Answer:

The coefficient of kinetic friction is #=0.16#

Explanation:

The mass of the object is #m=8kg#

The acceleration due to gravity is #g=9.8ms^-2#

The Normal reaction is

#N=mg=8xx9.8=78.4N#

The initial velocity is #u=16ms^-1#

The final velocity is #v=0ms^-1#

The distance is #s=80m#

Apply the equation of motion,

#v^2=u^2+2as#

The acceleration is

#a=(v^2-u^2)/(2s)=(0^2-16^2)/(2*80)=-1.6ms^-2#

According to Newton's Second Law

#F=ma#

The force of friction is

#F_r=8*1.6=12.8N#

The coefficient of kinetic friction is

#mu_k=F_r/N=12.8/78.4=0.16#