Blocks with masses of 3.0 kg, 4.0 kg, and 5.0 kg are lined up in a...

Blocks with masses of 3.0 kg, 4.0 kg, and 5.0 kg are lined up in a row on a frictionless table. All three are pushed forward by a 19N force applied to the 3.0 kg block.

a) How much force does the 4.0 kg block exert on the 5.0 kg block?

b)How much force does the 4.0 kg block exert on the 3.0 kg block?

please explain! thanks!

Expert Solution

Net mass m = m₁ + m₂ + m₃

= 3.0 + 4.0 + 5.0

= 12 kg

net acceleration a = F / m = 19.0 / 12.0

= 1.583 m/s²

a. Force exerted on 5 kg block F = m3 * a

= 5.0 * 1.583

= 7.915 N

b. Force exerted on 3 kg block F = (m₂+ m₃) * a

= (4.0 + 5.0) * 1.583

= 14.247 N

genius_generous answered 2 months ago

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