Question

Two stacked blocks are pushed along a horizontal table. Force F' is applied horizontally to the 6m block as shown: 5m 6m Assume there is kinetic friction at the interface between the 6m block and the table, and static friction at the interface between the two blocks. If m = 2 kg, ?s = 0.61, and ?k = 0.36, what is the maximum force F' that can be exerted without the top block slipping? Express your answer in N, to at least one digit after the decimal point.

          Two stacked blocks are pushed along a horizontal table.
Force F' is applied horizontally to the 6m block as shown:
5m
6m
Assume there is kinetic friction at the interface between the 6m block and the table, and static friction at the interface between the two blocks.
If m = 2 kg, ?s = 0.61, and ?k = 0.36, what is the maximum force F' that can be exerted without the top block slipping?
Express your answer in N, to at least one digit after the decimal point.
        
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Two stacked blocks are pushed along a horizontal table.
Force F' is applied horizontally to the 6m block as shown:
5m
6m
Assume there is kinetic friction at the interface between the 6m block and the table, and static friction at the interface between the two blocks.
If m = 2 kg, ?s = 0.61, and ?k = 0.36, what is the maximum force F' that can be exerted without the top block slipping?
Express your answer in N, to at least one digit after the decimal point.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Two stacked blocks are pushed along a horizontal table. Force F' is applied horizontally to the 6m block as shown: 5m 6m Assume there is kinetic friction at the interface between the 6m block and the table, and static friction at the interface between the two blocks. If m = 2 kg, μs = 0.61, and μk = 0.36, what is the maximum force F' that can be exerted without the top block slipping? Express your answer in N, to at least one digit after the decimal point.
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Transcript

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00:01 We'd like to find the max force f prime that can be exerted without the top block slipping.
00:07 So we want to use our newton's second law to find our equations of motion.
00:13 Our max force f prime is mu k times our sum of our masses, so mass 1 plus mass 2 times gravity, plus mu static times mass 1 times gravity.
00:28 So we can now substitute our values into this expression.
00:31 We can rewrite this as mu k times our two different blocks, so 5 plus 6 times g plus mu s times 5 times g.
00:49 Now we can plug in our values.
00:51 We have mu k, which is 0 .34 times we now want to substitute our values into this expression.
01:09 So, and this is, this is just, let's just say mass 1 plus mass 2...
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