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.
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
kg m/s^2 is obtained from F = m a, and is not equivalent to work in Joules, but to Newtons of force.
This force is the force of gravity, or the weight of the 64 kg object.
The 4.7 m is the distance through which the object is raised, so its gravitational potential energy increases by amount 4.7 m * 627.2 Newtons = 2800 Joules or so.
2.77 meters is the distance the spring is stretched.
A spring exerts a force F = - k x back toward the equilibrium position, where x is the distance of the stretch. So we know that
-876.4 Newtons = - k * 2.77 meters, which we solve for the force constant k, obtaining
k = 330 Newtons / meter, approximately.
A stretched spring has potential energy PE = 1/2 k x^2. For this spring you therefore get PE = 1200 Joules, very approximately.
Why do I not add the 876.4 N to the 627.2 J to get the KE?
You have the right idea about adding energies. However these quantities are not both energies.
N and J don't add. One measures force, the other measured energy. They are not like terms. You wouldn't add these any more than you would add, say, 10 miles to 40 feet.
Why do you multiply the 627.2J by the 4.7m?
it's 627.2 N, not 627.2 J. You multiply the weight by the distance it is raised against gravity to get the work done against gravity, which is the change in gravitational potential energy.
I would think I should take 876.4N / 64kg = 13.7m/s^2 to get the acceleration.
This would be the acceleration of a 64 kg mass on which a spring is exerting an unresisted force of 867.4 N. However note that as the spring contracts, the force changes, so the acceleration won't remain constant. You therefore can't use the equations of uniformly accelerated motion to analyze this situation.
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.
Conservative forces can indeed be experienced 'in reverse' when the object is returned to its original position.
However the key here is that energy, the work done against the conservative force, is returned.
Note that 876 N is a force, not an energy.
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.
You need to take the questions one at a time. First understand the situation, then build your understanding, one step at a time.
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 calculate an area corresponding to a region beneath a curve, you multiply the width of the region by its average height. It is often a good idea to break the region into trapezoids, and find the areas one at a time.
For example consider the region between the 4 cm and 12 cm stretches.
This region extends in width from 4 cm to 12 cm, a width of 8 cm.
Its 'heights' are 19.7 N, 32.5 N, 43.5 N; average 'height' appears to be a bit over 30 N.
Area would therefore be a bit over 8 cm * 30 N = 240 N * cm or 2.40 N * m.
The region applicable to the problem is from 0 to 11 cm; a good graph will yield area about 2.6 N * m or 2.6 J.
To get velocity I first find the acceleration: Do I use the 2.6203J as the force?
2.6 J is the energy stored in the system, not a force.
If this energy is returned to the system, then the system will have 2.6 J of kinetic energy.
Kinetic energy is 1/2 m v^2. So you set 1.2 m v^2 = KE and solve for v, obtaining v = +-sqrt( 2 KE / m).
Substituting the 'returned' 2.6 J for KE, you obtain the an expression for the velocity.
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?
If there is friction, then on a level surface coefficient of friction * weight = frictional force. In this case
f = frictional force = .07 * (.14 kg * 9.8 m/s^2) = .1 Newton, approx..
This force acts in a direction opposite to the motion of the object. So if the object slides thru distance `ds, displacement and the frictional force are in opposite directions and the work done by friction is negative; we get
`dW_frict_ON = - f * `ds,
where f represents the .1 N frictional force and `ds the unknown distance of the slide.
Intuitively, when the frictional force has dissipated the 2.6 J of KE, the object will come to rest. A force of .1 N can dissipate 2.6 J if it acts thru a distance of 26 meters.
More formally, starting with the instant the rubber band has transferred its PE to the KE of the system:
Since the object is given an initial KE of 2.6 J, and in the process of sliding to a stop the object reaches a state of 0 kinetic energy, its change in kinetic energy is -2.6 J.
The frictional force is nonconservative, and is the only nonconservative force acting on the object.
Since `dKE + `dPE = `dW_noncons_ON, and since gravitational PE is not changing on a level surface (recall that the rubber band is finished at the beginning of this phase, and its PE is no longer changin), we have
`dKE = `dW_noncons_ON.
So
-2.6 J = - .1 N * `ds and
`ds = -2.6 J / (.1 N) = 26 meters.
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.
The object will travel an unknown distance `ds along the incline. In the process its altitude will increase by `ds * cos(4.6 degrees), so its gravitational PE will increase by
`dPE_grav = .14 kg * 9.8 m/s^2 * (`ds * cos(4.6 deg)).
Friction will do work -f * `ds = -.1 N * `ds.
So
`dKE + `dPE = `dW_noncons_ON becomes
-2.6 Joules + .14 kg * 9.8 m/s^2 * (`ds * cos(4.6 deg)) = -.1 N * `ds.
Easier to do this in symbols:
`dKE + m g ( `ds cos(theta) ) = - f * `ds.
Solve for `ds. If you do the steps correctly you get
`ds = -`dKE / (m g cos(theta) + f).
Substitute for `dKE, m, g, theta and f and you get the distance along the incline.
Note that because of the incline, the force between object and incline will be a bit less than before, making the frictional force slightly less. For a 4.6 deg incline the difference is slight, and has been neglected here.
My responses will hopefully help clarify some important ideas; they are also likely to raise as many questions as they answer; feel free to ask them.