Question

Two forces of 6n and 3n are acting on the two blocks of 2kg and 1kg kept on frictionless floor. What is the force exerted on 2kg block by 1kg block?

Answer 1

Answer: 4 N

Explanation:

F.B.D of 2 kg block = 6 - F = 2a, F.B.D of 1 kg block = F - 3 = a, putting value a in first equation 6 - F = 2F - 6 = 6 + 6 = 2F = F = 12/3 = 4 N

Answer 2

Answer:

Let F be force exerted on 2kg block by 1kg block,

So Free body diagram:

For 2kg block,

6−F=2×a−−−(1)

where a is the acceleration of two blocks

For 1kg block

F−3=1×a⇒a=F−3

Put this value in equation (1)

6−F=2×(F−3)⇒3F=12⇒F=4N

Explanation:

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