Force of Friction: 0.45N, Coefficient: 0.0382

[italystaly]

A force of 1.2N is applied to an object of mass 1.5Kg. It accelerates at 0.50 m/s^2. Determine the Force of friction and the Coefficient of Friction

This is what I did. However, it seems that the Force of Friction is really low so I'm not sure if I did this question right.
F=M * A ..which is can also be used as Fnet
.:. Fnet= 1.5kg(0.50 m/s^2)
= 0.75 N

Fapp= 1.2 N

Fnet=Fapplied + Ffriction
Ffriction = Fnet-Fapplied
= 0.75n -1.2 n
= - 0.45 N


If I continue to Figure out the coefficient of Friction it seems very unreasonable.

Ffriction/ Fnormal = kinetic Coefficient

-.45n/ -9.8m/s^2 * 1.5kg
= .0382



This may be right but I'm not quite sure. Any input would be great.

 

[kuruman]

Draw a free body diagram. Say the applied force is to the right. Then the force of friction must be to the left. So from the FBD you get

Fnet=Fapplied - Ffriction, not what you have.