How do you find the acceleration of two objects if the coefficient of kinetic friction between the 7kg object and the plane is 0.250? Given: m1=7kg, m2=12kg, and there is an angle of 37.0 degrees.
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I need more information. What does the set-up look like?
Okay. There is a triangle, with a 37 degree angle in the left bottom corner, holding up a block that weighs 7kg. This block is attached to a levy on the right upper corner of the triangle which is holding up a 12 kg block hanging downward on the right.
First draw a free body diagram. For the 7kg object you should get the two equations: Fx=T-Ff-mgsin37=m1a and Fy=N-mgcos37=0. So N=mgcos37=54.79. The friction force(Ff)=.25N=13.7. Substitute this into Fx and you will get Fx=T-54.98=m1a. This is the same as T=m1a+54.98. Then from the 12kg object, you should get the equation: Fy=m2g-T=m2a. Which is the same as T=m2g-m2a. Substitute this in for the T in the Fx equation and solve for a. I got 2.76, but I may have done some math wrong somewhere, so you should probably check everything :)