Question

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 16 N force applied to the 3.0 kg block. How much force does the 4.0 kg block exert on the 5.0 kg block?

Answer 1

6.7 N

Step-by-step explanation:

The force exerted by the 4.0 kg block on the 5.0 kg block will be 6.7

Fnet = 16 N (since it is applied only to the 3.0 kg block)

Total mass of the system, m = 3.0 kg + 4.0 kg + 5.0 kg = 12.0 kg

Fnet = ma16 = 12aSolving for a,a = 16/12 = 4/3 m/s²

m is the mass of the 5.0 kg block and a is the acceleration of the system.F = 5.0 kg x 4/3 m/s² = 6.666 N

Therefore, the force exerted by the 4.0 kg block on the 5.0 kg block is 6.7 N.

Answer 2

2019?

Step-by-step explanation: