A 3.0-kg block sits on top of a 5.0-kg block which is on a horizontal surface. T

Question

A 3.0-kg block sits on top of a 5.0-kg block which is on a horizontal surface. The 5.0-kg block is pulled to the right with a force F?  as shown in the figure(Figure 1) . The coefficient of static friction between all surfaces is 0.64 and the kinetic coefficient is 0.44.

Part A.

What is the minimum value of F needed to move the two blocks?

Express your answer using two significant figures.

Part B.

If the force is 10% greater than your answer for (a), what is the acceleration of each block?

Express your answer using two significant figures.

Explanation / Answer

N1=mg
f1=N1u=mgu
T-f1=ma
N2=(m+M)g
f2=(m+M)gu
F-T-f1-f2=Ma

1. F-f1-f1-f2=0
F-2f1-f2=0
F=2mgu(s)+(m+M)gu(s)
=u(s)(3m+M)g
=(.64)(14kg)(9.80m/s^2)
=87.808 N

2. F = 1.1*87.808= 96.588
a=[F-(3m+M)gu(k)]/(m+M)
=[96.588-(.44)(14 kg)(9.80 m/s^2)]/(8 kg)
=4.5275m/s^2

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