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October 24, 2014

October 24, 2014

Posted by **Sandhya** on Sunday, September 21, 2008 at 5:31am.

(a) Determine the acceleration of the system

(b) Determine the tension in the cord connecting the 3.0 kg and the 1.0 kg blocks.

(c) Determine the magnitude of the force exerted by the 1.0 kg block on the 2.0 kg block.

- college physics -
**drwls**, Sunday, September 21, 2008 at 8:13ama) The applied force of F = 46N acts on the system of all three blocks. You can ignore the internal foces (cord and contact) between blocks and apply F = (M1 + M2 + M3) a, where M1 = 1 kg, M2 = 2 kg and M3 = 3 kg.

a = 46 N/6 kg = 23/3 m/s^2

b) Let f1 be the cord tension. Only F and f1 act on M3, the 3 kg mass. Apply Newton's law.

F - f1 = M3 * a

Solve for f1. Use the value of a from part (a)

(c) The only force applied to M2 is the contact force from M1. Call it f2. Solve

f2 = M2 a

- college physics -
**Sandhya**, Sunday, September 21, 2008 at 12:28pmThanks a lot for your timely help.

- college physics -

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