I found acceleration and the net force acting on the 4.6 kg block. How do I find
ID: 1331494 • Letter: I
Question
I found acceleration and the net force acting on the 4.6 kg block. How do I find the force from the 6.9 kg block on the 4.6 kg block?
011 10.0 points Three blocks in contact with each other are pushed across a rough horizontal surface by a 100 N force as shown. The acceleration of gravity is 9.8 m/s | | 2 kg | 4.6 kg || 6.9 kg = 0.15 If the coefficient of kinetic friction between each of the blocks and the surface is 0.15, find the magnitude of the force exerted on the 4.6 kg block by the 6.9 kg block. Answer in units of N.Explanation / Answer
Force exerted on 4.6kg block by 6.9kg block = µmg = 0.15x6.9x9.8 = 10.183 N
As I understand, you are asking about the force exerted by 6.9 kg block on 4.6 kg block. The net force shall be the frictional force which we have calculated above. If you have any further doubt then please write me.
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