course PHY 121
I will take my test on Monday. I have a question on the problem sets. I went back and attempted all the problems in set 3 and 4 again and I did a bit better.
Problem Set 3 Number 17A mass of 64kg is raised 4.7meters and at the same time stretches an ideal spring 2.77meters from its equilibrium length, at which point the spring exerts a force of 876.4 Newtons.
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If the system is then released from rest, how much KE will it have when it reaches its original position and what will be its velocity?
How does this example illustrate the nature of a conservative force?
m = 64kg f=876.4N I don't know what the 2.77m is and I assume the 4.7m is ds vo = 0
I know I take the mass * g = 64kg * 9.8m/s^2 = 627.2kg m/s^2 or Joules
Why do I not add the 876.4 N to the 627.2 J to get the KE?
Why do you multiply the 627.2J by the 4.7m?
I would think I should take 876.4N / 64kg = 13.7m/s^2 to get the acceleration. From here I would take vf^2 = vo^2 + 2a ds
vf^2 = 0 + 2(13.7m/s^2)(4.7m)
vf^2 = 128.78?
vf= sqrt of 128.78 = 11.35m/s
therefore the change in velocity is 11.35m/s and the average velocity is 11.35m/s / 2 = 5.67m/s
This example represents a conservative force because the force is returned to the system. If it exerts 876N this 876N will be returned to the object when it comes to rest back at its original position, barring other forces intervening.
I don't understand how the problem set is getting their answer, or more than that why.
Problem Set 3 Number 14
A rubber band exerts forces of 19.7N, 32.5N, and 43.5N when stretched by 4cm, 8cm, and 12 cm.
If the rubber band is stretched a distance of 11cm and used to accelerate a mass of .14kg, what velocity will the mass attain if there is no friction acting on the mass?
If the coefficient of friction between the mass and the horizontal surface over which it travels is .07, then how far will the object travel before coming to rest?
If the mass travels without friction up a ramp inclined at angle 4.6degrees with horizontal, how far up the incline will it travel? How far along the incline will it travel?
These are the problems that cause trouble? I get overwhelmed with all the questions.
The problem says the work to stretch(area under the curve) is 2.6203J. I do not see how this was gotten? Do you subtract two points on the y axis and two points on the x axis to get the rise over run or the slope.
To get velocity I first find the acceleration: Do I use the 2.6203J as the force? 2.6203J/.14kg = 18.7cm/s^2 I then use vf^2 = vo^2 + 2a ds
vf^2 = 0 + 2(18.7cm/s^2)(11cm) = vf^2 = 411.4cm/s^2 vf = 20.28cm/s
thus the dv = 20.28cm/s and the vAve = 20.28cm/s / 2 = 10.14cm/s
I do not understand about the coefficient, is there an easier way to explain what this is and how it is used than what I have heard thus far? Can it be related to a real life situation so that I understand it?
When the mass climbs up the incline I multiply the mass of .14kg * 9.8m/s^2 and get 1.37N of force. I do not understand where to go from here.
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Good. Let me know if you have questions.