If a 1 kg object moving at 10 m/s slows down to a halt after moving 50 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

1 Answer
Dec 8, 2015

mu=1/9.8

Explanation:

First we must find the acceleration experienced by the object.
Here we will use the kinematic equation:
(vf)^2=(vi)^2+2ad
plug in our given data...
0=100+100a
a=-1m/s^2
We can now use this acceleration to calculate force using F=ma
F=1kg*-1m/s^2
F=-1N this must be the force of friction because no other resistive forces are given.

Kinetic friction is fk=muN
The normal force in this case is the weight of the object. N=mg
N=1kg*9.8m/s^2
N=9.8

Plugging in our values for fk and N....
1=mu*9.8
mu=1/9.8