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November 27, 2014

November 27, 2014

Posted by **ericka** on Monday, May 7, 2012 at 10:28pm.

Question # 2 : A 1.5-kg block rests on top of a 7.5kg block. The cord and pulley have negligible mass, and there is no significant friction anywhere.

Part a) What force F must be applied to the bottom block so the top block accelerates to the right at 2.9 m/s^2 ?

Part b) What is the tension in the connecting cord?

- physics -
**Elena**, Tuesday, May 8, 2012 at 11:36amQ1:

The car moves forward by the reaction force from the ground, produced due to friction between tires and road. According to the Newton’s 3rd law

F12 = F21

F(net)max = F(friction)max

m•a = k• m•g

a = k• g = 0.82•9.8 = 8.04 m/s^2

Q2:

m1 =1.5 kg, m2 = 7.5 kg, a=2.9 m/s²

(a) The horizontal projections of the equations of motion for each block are

m1•a = T,

m2•a = T-F,

F = (m1+m2) •a = (1.5+7.5) •2.9 = 26.1 N,

(b) T= m2•a - F= 7.5•2.9 – 26.1 = 4.35 N.

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