Question #156458

As shown in the figure below, two forces are applied horizontally to two blocks in contact. The horizontal surface on which the blocks slide frictionless. if F = 6.0N and M=1.0kg, what is the magnitude of the force exerted on the large block of mass 3M by the small block of mass 2M.

https://i.gyazo.com/25f5ef792c93e6b6a36a11cfb984d5a5.png


1
Expert's answer
2021-01-19T07:08:27-0500


Let's first find the acceleration of the system of two blocks from the Newton's Second Law of Motion:


2FF=(3M+2M)a,2F-F=(3M+2M)a,a=F5M=6.0 N51.0 kg=1.2 ms2.a=\dfrac{F}{5M}=\dfrac{6.0\ N}{5\cdot1.0\ kg}=1.2\ \dfrac{m}{s^2}.

Then, we can find the net force on the first block with mass 3M3M:


F1,net=3Ma=31.0 kg1.2 ms2=3.6 N.F_{1,net}=3Ma=3\cdot1.0\ kg\cdot1.2\ \dfrac{m}{s^2}=3.6\ N.

Finally, we can find the magnitude of the force exerted on the large block of mass 3M by the small block of mass 2M:


2FF21=F1,net,2F-F_{21}=F_{1,net},F21=2FF1net=26.0 N3.6 N=8.4 N.F_{21}=2F-F_{1net}=2\cdot6.0\ N-3.6\ N=8.4\ N.

Answer:

F21=8.4 N.F_{21}=8.4\ N.


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