If a #2 kg# object moving at #9 m/s# slows down to a halt after moving #2 m#, what is the friction coefficient of the surface that the object was moving over?

2 Answers
Aug 29, 2016

By conservation of energy the initial KE of the object must be same as the work done by it against the frictional force. So

#1/2mv^2=mumgxxx....(1)#

Where

# m -> "mass of the object"#

#mu->"coefficient of friction"#

#x->"displacement"#

#g->"Acceleration due to gravity"#

From (1)

#mu=v^2/(2xxgxxx)#

#mu=9^2/(2*9.8.2)=81/(4*9.8)=2.06#

Sep 12, 2016

Answer:

Alternate solution.

Explanation:

Using the kinematic equation
#v^2-u^2=2as# where symbols have respective usual meanings, we get
#0^2-9^2=2xxaxx2#
#a=-81/4ms^-2#
Decelerating force due to friction is #=ma#
#=2xx(-81/4)=-81/2N#

Equating it with Force due to friction #=-mumg#, we obtain
#81/2=muxx2xx9.81#
Solving for #mu# we get
#mu=(81/2)/(2xx9.81)=2.06#, rounded to two decimal places.