I am having trouble with a problem from Physics for Scientists and Engineers by Tipler and Mosca, 6th edition, 5.56
A 100-kg mass is pulled along a frictionless surface by a horizontal force F such that its acceleration is a1=6.00m/s^2. A 20.0-kg mass slides along the top of the 100-kg mass and has an cceleration of a2=4.00m/s^2. (It thus slides backward relative to the 100-kg mass. (a)What is the frictional force exerted by the 100-kg mass on the 20.0 kg mass? (B)What is the net force acting on the 100-kg mass? What is the force F? (C) After the 20.0-kg mass falls off the 100-kg mass, what is the acceleration of the 100-kg mass? (Assume that the force F does not change).
For my free body diagram, postive x and y are to the right and up, respectively.
I attempted to solve this problem using relative acceleration and setting a=0.
a1=6.00, a2=4.00. Therefore, ar=1.00m/s^2 (?)
The Attempt at a Solution
(a) Fnetm2y=Fn- m2 g=0
-fk=-[tex]\mu[/tex]k m2 g=m2a2
And fk=(.1019)(20kg)(9.81m/s^2)=20.0N (?)
Force F=(m1+m2)ar=(120kg)(1.00m/s^2=120N (?)
(c) F=m ar when m2 falls off
I am not sure about my answers because I don't know whether or not I used the relative acceleration concept correctly. Are my answers correct? Is this the only way to do the problem? Is there a way that it can be solved without using relative accelerations?
I googled this problem and found a site that solved it very differently and got different answers. I am not allowed to insert the link here. It did not use relative accelerations.
Any comments and/or answers would be very much appreciated. Thanks!!