One of the most difficult parts in computing work of a force is determining the SIGN of the work. My suggestion on this (and all problems for which you need to calculate work) is to start with the definition of work:
U_12 = integral(F • e_t ds)
where F is the force VECTOR, e_t is the unit VECTOR of the point at which the point acts, ds is the differential path length and "•" is the dot product operation.
- For part a), the spring force can be written as -k*x*i. The motion is in the +x direction (e_t = +i) and ds = dx. Using the above equation for work, we have: U_12 = -k*integral(xi • i dx) from x = -6 to x = +3.
- For part b), the weight force vector is: W = mg*(-sin(15°)*i - cos(15°)*j). As above, e_t = +i and ds = dx. Using the above equation for work, we have: U_12 = mg*integral((-sin(15°)*i - cos(15°)*j) • i dx) from x = -6 to x = +3. (Note that since the x-component of weight is always in the -x direction and motion is in the +x direction, the work due to the weight force will be NEGATIVE. Use this observation to check your answer.)
If you use this approach, the signs will take care of themselves -- you will NOT need to do any additional thinking about the sign.
Although the path in this problem is pretty simple, it is still good to go through the formality of the above operation. What you learn here will help you in problems that are more difficult to visualize the motion.
Let me know if you do not know how to do this operation.
7 comments:
Sorry, I had a mistake in my equation and didn't want to confuse anyone. Anyhow, I think I got the answer on this question but I was wondering if I found k the correct way? I assumed that since it was at rest the force of the spring had to balance the force of the weight.
-k*s = -W * sin(theta)
Is that method correct?
Not sure of your question. You are GIVEN the spring constant k for the problem: k = 3 lb/in (or 36 lb/ft).
The equation that you give is correct for finding the static equilibrium position of the block in the absence of the force P (however, P would affect this static deformation). Keep in mind that the block is not in equilibrium.
Never mind, I overlooked the k value in the drawing.
So we are basically just integrating kx*dx on the appropriate interval?
I am unsure of what to do with the i's in the integral.
Please note that i•i = 1 and i•j = 0.
a) With the above, the integral becomes:
integral(x*i•i dx) =integral(x dx)
b) With the above, the integral becomes:
integral(-sin(15°) dx)
is mass in this 14, or 14/32.2?
thanks
mass is 14/g = 14/32.2
so mg = 14
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