A block of mass 2.69 kg is placed on top of a second block of mass 5.13 kg. The

Question

A block of mass 2.69 kg is placed on top of a
second block of mass 5.13 kg. The coefficient
of kinetic friction between the 5.13 kg block
and the surface is 0.29. A horizontal force F
is applied to the 5.13 kg block.Calculate the magnitude of the force neces-
sary to pull both blocks to the right with an
acceleration of 1.73 m/s2. Assume no slipping
between the two blocks. The acceleration of
gravity is 9.8 m/s2 .Answer in units of N

Explanation / Answer

2.69+5.13=7.82kg 7.82x9.8 = N= 76.636N ======> friction = 76.636Nx0.29= 22.22N ma = F - friction 7.82x1.73 + 22.22 = F = 35.75N

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