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course Phy 241
023. Forces (atwood, chains)
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Question: `q001. Note that this assignment contains 3 questions.
A chain 200 cm long has a density of 15 g/cm. Part of the chain lies on a tabletop, with which it has a coefficient of friction equal to .10. The other part of the chain hangs over the edge of the tabletop.
If 50 cm of chain hang over the edge of the tabletop, what will be the acceleration of the chain?
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Your solution:
50 cm * (15 g/cm)
= 750 g = .75 kg
9.8 m/s ^ 2 * .75 kg
= 7.3 Newtons
.10 * 21 Newtons
= 2.1 Newtons
5.2 N / (3 kg)
= 1.7 m/s^2
confidence rating #$&*:
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Self-critique (if necessary):
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Self-critique rating:
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Question: `q002. What is the maximum length of chain that can hang over the edge before the chain begins accelerating?
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Your solution:
(x * .015 kg / cm) * 9.8 m/s^2
= x * .147 N / cm
(200 cm - x) * .015 kg / cm * 9.8 meters/second ^ 2
= .147 N / cm * (200 cm - x)
.0147 N / cm * (200 cm - x) = .147 N/cm * x
200 cm - x = 10 * x
200 cm = 11 x
x = 200 cm / 11
x = 18 cm
confidence rating #$&*:
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Self-critique (if necessary):
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Self-critique rating:
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Question: `q003. The air resistance encountered by a certain falling object of mass 5 kg is given in Newtons by the formula F = .125 v^2, where the force F is in Newtons when the velocity v is in meters/second. As the object falls its velocity increases, and keeps increasing as it approaches its terminal velocity at which the net force on the falling object is zero, which by Newton's Second Law results in zero acceleration and hence in constant velocity. What is the terminal velocity of this object?
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Your solution:
5 kg * 9.8 meters/second ^ 2
= 49 Newtons
v = sqrt(49 Newtons/(.125 ))
v = `sqrt (392 Newtons)
v = 19.8 m/s
confidence rating #$&*:
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Self-critique (if necessary):Dont know exactly how to cancel the given unit to produce a m/s for my answer.
@&
The rule for F would be written, in terms of units, as
F = .125 N / (m/s)^2 * v^2
N / (m/s)^2 could also be reduced to fundamental units of kg, m and s.
Either way, the units of your calculation would then work out.
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&#This looks good. See my notes. Let me know if you have any questions. &#