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Fall 2008 -- Purdue University -- West Lafayette, IN
Blog Post LABELS
- Blog Challenge Questions (9)
- Chapter 2 (2)
- Chapter 3 (19)
- Chapter 5 (11)
- Chapter 6 (23)
- Chapter 7 (7)
- Chapter 8 (9)
- Exam 1 (5)
- Exam 2 (4)
- Exam 3 (17)
- Exams (2)
- Final Exam (7)
- General (6)
- Gravity (2)
- Homework (2)
- in class (1)
- Lectures (1)
- Quizzes (2)
- Spring 2008 Archive (87)
- Spring-Mass Simulator (1)
- Student-Generated Solutions (1)
- Summer 2008 Archive (72)
11 comments:
Is it correct when finding the spring work to just use the equation (k/2) * (l - l_0)^2 ? Essentially i am asking can we ignore everything else and still get the correct work of just the spring using this equation.
I am slightly confused as to the question, is it asking the work done by the spring from x1 through x2 and then the work sone by the wieght from the neutral postition to x1, at which case one could then jsut look at the spring vs distance graph and compute the area under the graph, which is the same as using the equation.
i think the question simply ask the workd done by spring force (k/2 (x2-x1)) and work done by gravity (mg (h2-h1)).
I am also slightly confused as it say the spring is unstretched at x=0 and the x-axis starts from the end of the unstretched spring. However, it says that the initial position is x1=-6in. Does that mean that the cart is pushed in the -x direction and then let go such that it reaches x2=3. Thus travelling a total of 9 inches?
Let me jump in to talk about what is being asked in this problem and about how you might go about working the problem.
* The problem states that the spring is unstretched when x = 0. All that this does is to define from what point x is measured. That is, when the spring is compressed, x < 0; when the spring is stretched, x > 0.
* The spring starts out at x = -6 in. That is, the spring is compressed by 6 in at the beginning. The spring ends up at x = 3 in; that is, the spring ends up being stretched by 3 in.
* There are three forces that do work on the system: P, the spring force and the gravitational force. Therefore, we can break up the work done on the system into three parts: U_total = U_P + U_spring + U_grav.
* You are asked to find two things: U_spring and U_grav.
One way to work this is to find the work done by the spring or gravity by integration (using the definition of work). However, since both the spring and gravitational forces are conservative, we can find the work by these forces as:
U_spring = -(V_2 - V_1)_spring
U_grav = -(V_2 - V_1)_grav
The latter is recommended.
Let us know if this does not help.
If you set your datum line on the bottom of the box in the picture given (x=0) wouldn't your work done by v_grav just be V_2? Or would it be better to place your datum line on the top of the box at position 1?
So...
if instead the spring moved from
x = -3 to x = 3,
the work done by the spring would be zero..
and work is negative if force is in opp dir of movement of object and positive if in same dir..
does it make sense to have negative work?
--to Brian--
If you place the datum at x = -6 in, then V1 = 0 and therefore, U_grav = -V2. If you place the datum at x = 0, you have non-zero values for both V1 and V2.
I do not see any real advantage of one datum line placement over any others in this problem.
--to ClayDT3--
Yes, in moving from x = -3 in to x = +3 in, the spring does no net work.
--to wccheng--
Negative work occurs when a force acts in a way that it opposes the motion of the body on which it acts. The negative sign is important since this tells you that the force in that case tends to decrease the speed of the body.
This was the case on Problem No. 2 of Quiz 7: in that problem, friction opposed the motion of the block and will tend to decrease its speed.
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