If a #2 kg# object moving at #5 m/s# slows down to a halt after moving #100 m#, what is the friction coefficient of the surface that the object was moving over?

1 Answer
Jan 24, 2018

Answer:

#0.0125#

Explanation:

Here,if constant retardation occurring due to frictional force is a then we can use the equation #v^2=u^2-2as#(all symbols are bearing their conventional meaning) where, #v=0# ,#u=5m/s# and #s=100#

So, #a = 0.125 m/s^2#

So,here amount of force acting against the motion of the object is #m×a# or #(2×0.125) N#

And this will be equal to the amount of frictional force acting at their interface i.e #mu×N# = #mu×mg# (where, #mu#= coefficient of frictional force)

So,we can write,

#mu×2×10 = 2 × 0.125# or #mu# = #0.0125#