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

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Blocks with masses of 3.0 kg, 4.0 kg, and 5.0 kg are lined up ina row on a frictionless table. All three are pushed forward by a19N force applied to the 3.0 kg block.

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

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

please explain! thanks!

Answer & Explanation Solved by verified expert
4.1 Ratings (770 Votes)

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


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