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December 20, 2014

December 20, 2014

Posted by **QUESTION** on Thursday, August 20, 2009 at 6:10pm.

A 4.0 kg block is stacked on top of a 12.0 kg block, whcih is accelerating along a horizontal table at a = 5.2 s^-2 m. Let Mu k = Mu s = Mu.

(a) What mininum coefficent of friction Mu between the two blocks will prevent the 4.0 kg block from sliding off?

(b) If Mu is only half this mininium vale, what is teh acceleration of the 4.0 kg block with respect to the table, and

(c) with respect to the 12.0 kg block?

(d) What is teh force taht must be applied to the 12.0 kg block in (a) and in (b), assuming that the table is frictionless?

for (a) I got .66

ok for (b) I got 3.2 s^-2 m

for (c) I got 2.0 s^-2 m

And I don't understand what the problem means by

"(d) What is the force taht must be applied to the 12.0 kg block in (a) and in (b), assuming that the table is frictionless?"

Does this mean that for (a) the bottom block experiances a force of kinetic friction were Mu is the same Mu i found in part(a) of .66????

part (b) is frictionless with the table... I know how to do that just don't understand what the other part is asking??????

- Physics -
**QUESTION**, Thursday, August 20, 2009 at 6:12pmSense this problem 32 the answer isn't in the back =[

- Physics -
**QUESTION**, Thursday, August 20, 2009 at 6:53pmcan you check my answer for (c)

- Physics -
**QUESTION**, Thursday, August 20, 2009 at 6:59pmok for the applied force for (a) on the bottom block I got 83 N not sure if it's right...

for the applied force in (b) wouldn't it be the same... I'm confused...

Can you check my answer for (c)

- Physics -
**bobpursley**, Thursday, August 20, 2009 at 8:38pmThe bottom block has no friction with the table.

If you know the acceleration, then the force applied must be (totalmass)a where totalmass is 12+4 kg.

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