Phy 231
Your 'cq_1_15.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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A rubber band begins exerting a tension force when its length is 8 cm. As it is stretched to a length of 10 cm its tension increases with length, more or less steadily, until at the 10 cm length the tension is 3 Newtons.
• Between the 8 cm and 10 cm length, what are the minimum and maximum tensions?
answer/question/discussion: ->->->->->->->->->->->-> :
The minimum is at 8 cm and is 0 N.
The maximum is at 10 cm and is 3 N.
#$&*
• Assuming that the tension in the rubber band is 100% conservative (which is not actually the case) what is its elastic potential energy at the 10 cm length?
answer/question/discussion: ->->->->->->->->->->->-> :
W = average force * ds
average force = (3 N + 0 N)/2 = 1.5 N
W = 1.5 N * 2 cm
w = 3 Ncm
3 Ncm * 0.01 m = 0.03 J
#$&*
• If all this potential energy is transferred to the kinetic energy of an initially stationary 20 gram domino, what will be the velocity of the domino?
answer/question/discussion: ->->->->->->->->->->->-> :
KE = ½*m*v^2
20 g = 0.02 kg
0.03 J = ½ *0.02 kg * v^2
v = 1.73 m/s
#$&*
• If instead the rubber band is used to 'shoot' the domino straight upward, then how high will it rise?
answer/question/discussion: ->->->->->->->->->->->-> :
m*g*h= 0.03 J
h= 0.03 J /(0.02 kg * 9.8 m/s/s)
h = 0.153 m
#$&*
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45 mins
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&#Very good work. Let me know if you have questions. &#
Phy 231
Your 'cq_1_15.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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015. Impulse-Momentum
Question: `q001. Note that this assignment contains 5 questions.
. Suppose that a net force of 10 Newtons acts on a 2 kg mass for 3 seconds. By how much will the velocity of the mass change during these three seconds?
Your solution:
Fnet = m*a
a = Fnet/m
a = 10 N/2kg
a = 5 m/s/s
5 m/s/s * 3 sec = 15 m/s
Confidence Assessment:3
Given Solution:
The acceleration of the object will be
accel = net force / mass = 10 Newtons / (2 kg) = 5 m/s^2.
In 3 seconds this implies a change of velocity
`dv = 5 m/s^2 * 3 s = 15 meters/second.
Self-critique (if necessary):
ok
Self-critique Assessment:3
Question: `q002. By how much did the quantity m * v change during these three seconds?
What is the product Fnet * `dt of the net force and the time interval during which it acted?
How do these two quantities compare?
Your solution:
M = 2kg
m*v = 2kg * 15 m/s = 30 N
Fnet * dt = 10 N * 3 sec = 30 N*s
Confidence Assessment:3
Given Solution:
Since m remained constant at 2 kg and v changed by `dv = 15 meters/second, it follows that m * v changed by 2 kg * 15 meters/second = 30 kg meters/second.
Fnet *`dt is 10 Newtons * 3 seconds = 30 Newton * seconds = 30 kg meters/second^2 * seconds = 30 kg meters/second.
The two quantities m * `dv and Fnet * `dt are identical.
Self-critique (if necessary):ok
Self-critique Assessment:3
Question: `q003. The quantity m * v is called the momentum of the object.
The quantity Fnet * `dt is called the impulse of the net force.
The Impulse-Momentum Theorem states that the change in the momentum of an object during a time interval `dt must be equal to the impulse of the average net force during that time interval. Note that it is possible for an impulse to be delivered to a changing mass, so that the change in momentum is not always simply m * `dv; however in non-calculus-based physics courses the effective changing mass will not be considered.
If an average net force of 2000 N is applied to a 1200 kg vehicle for 1.5 seconds, what will be the impulse of the force?
Your solution:
Force*time = impulse
2000 N * 1.5 sec = 3000 N *s
impulse = 3000 N*s
Confidence Assessment:3
Given Solution:
The impulse of the force will be Fnet * `dt = 2000 Newtons * 1.5 seconds = 3000 Newton*seconds = 3000 kg meters/second. Note that the 1200 kg mass has nothing to do with the magnitude of the impulse.STUDENT COMMENT: That's a little confusing. Would it work to take the answer I got of 3234 N and add back in the weight of the person at 647 N to get 3881?
INSTRUCTOR RESPONSE: Not a good idea, though it works in this case.
Net force = mass * acceleration.
That's where you need to start with problems of this nature.Then write an expression for the net force, which will typically include but not be limited to the force you are looking for. *&*&
Self-critique (if necessary):ok
Self-critique Assessment:3
Question: `q004. If an average net force of 2000 N is applied to a 1200 kg vehicle for 1.5 seconds, what will be change in the velocity of the vehicle?
Your solution:
Fnet = m*a
a = Fnet/m
a = 2000 N / 1200kg
a = 1.67 m/s/s
a = change in velocity/dt
1.67 m/s/s = change in velocity / 1.5 sec
change in velocity = 2.51 m/s
Confidence Assessment:3
Given Solution:
The impulse of the 2000 Newton force is equal to the change in the momentum of the vehicle. The impulse is
impulse = Fnet * `dt = 2000 Newtons * 1.5 seconds = 3000 Newton*seconds = 3000 kg meters/second.
The change in momentum is m * `dv = 1200 kg * `dv.
Thus
1200 kg * `dv = 3000 kg m/s, so
`dv = 3000 kg m/s / (1200 kg) = 2.5 m/s.
In symbols we have Fnet * `dt = m `dv so that
`dv = Fnet * `dt / m.
Self-critique (if necessary):ok
Self-critique Assessment:3
Question: `q005. Use the Impulse-Momentum Theorem to determine the average force required to change the velocity of a 1600 kg vehicle from 20 m/s to 25 m/s in 2 seconds.
Your solution:
Mass* change in velocity = momentum
change in velocity = 25 m/s- 20 m/s
change in velocity = 5 m/s
1600 kg *5 m/s = momentum
momentum = 8000 kg m/s
momentum = impulse
impulse = force * time
8000 kg m/s = f * 2 sec
f = 4000 N
Confidence Assessment:3
Given Solution:
The vehicle changes velocity by 5 meters/second so the change in its momentum is m * `dv = 1600 kg * 5 meters/second = 8000 kg meters/second. This change in momentum is equal to the impulse Fnet * `dt, so
Fnet * 2 sec = 8000 kg meters/second and so
Fnet = 8000 kg meters/second / (2 seconds) = 4000 kg meters/second^2 = 4000 Newtons.
In symbols we have Fnet * `dt = m * `dv so that Fnet = m * `dv / `dt = 1600 kg * 5 m/s / ( 2 s) = 4000 kg m/s^2 = 4000 N.
Self-critique (if necessary):ok
Self-critique Assessment:3
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35 mins
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