A horizontal (rightward) force of given magnitude F is exerted on the leftmost f

Question

A horizontal (rightward) force of given magnitude F is exerted on the leftmost face of two blocks sitting next to each other on a slippery surface, with two of their faces touching. The mass of the block on the far right is m, and the mass of the block on the left (the one that is being pushed) is 2m. Suppose instead an equal but oppositely directed force had been applied to the rightmost block, the one which had mass m. (In other words, now the external force F pushes the original system to the left), what would now be the magnitude of the contact force between the two blocks?

Explanation / Answer

The force equation for the right block will be:

F - T = ma ...................[1]

where T is the contact force between the two blocks.

and force equation for the left block will be:

T = (2m)a

=> T/2 = ma ..............[2]

substituting the 2nd equation in the first we get:

F - T = T/2

=> F = T + (T/2)

therefore F = (3T/2)

or simply rearranging it we get:

T = (2F/3) [1st option].

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