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From a previous problem we concluded that coefficient of kinetic friction is .34 and static is .43

I do not understand how to determine the blocks mass to be able to determine the Fpush needed. any tips

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- Thread starter Paulbird20
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- #1

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From a previous problem we concluded that coefficient of kinetic friction is .34 and static is .43

I do not understand how to determine the blocks mass to be able to determine the Fpush needed. any tips

- #2

cepheid

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I do not understand how to determine the blocks mass to be able to determine the Fpush needed. any tips

Neither do I.

Was this the exact statement of the problem? If not, can you please use the proper template for posting homework questions so that we are sure all of the pertinent information has been posted?

- #3

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A sufficiently large push is given to start the block moving. After that, what magnitude Fpush must be applied to make the block accelerate to the right at 5.71 m/s2

But i think the 7.4 is from a different question and is not related to this one or maybe it is?

- #4

cepheid

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If it is not a sub-part of a larger problem, then I find it very bizarre that a *stand-alone* problem would be worded in this way. What you have posted here is not enough info to determine Fpush.

- #5

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- #6

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The block is at rest on a table.

- #7

cepheid

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Soooo....can you now answer the problem?

- #8

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-fk = ma correct?

so 5.71x5 kg = 28.55 is the force required to accelerate it at 5.71

so 5.71x5 kg = 28.55 is the force required to accelerate it at 5.71

- #9

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and the table overall force is up (perpendicular to the table correct)?

- #10

cepheid

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-fk = ma correct?

Not really. The expression ma gives the NET force, which is the sum of two opposing forces, the applied force (which you are trying to calculate) and the frictional force (which you know).

so 5.71x5 kg = 28.55 is the force required to accelerate it at 5.71

You know that 28.55 N can't be the applied force, because if you subtract the frictional force from it to get the net force, and then divide this by 5kg, you don't get an acceleration of 5.71.

- #11

cepheid

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and the table overall force is up (perpendicular to the table correct)?

No, the table exerts two forces on the block. The overall force will be the sum of these.

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