- #1
omarMihilmy
- 31
- 0
My problem is with the solution
If we differentiate:
dU/dx = 9x^2 - 7
From where does the negative come from also the j component?
Potential Energy: Differentiating to Find Force
- Thread starter omarMihilmy
- Start date
AI Thread Summary
To find the force from potential energy, the correct approach is to differentiate the potential energy function U with respect to position x, yielding dU/dx = 9x^2 - 7. Concerns were raised about the presence of the negative term and the j component in the differentiation. The relevant equation for force derived from potential energy is F = -dU/dx, indicating that the force is the negative gradient of potential energy. Additionally, it was noted that the y-component should not be disregarded during differentiation. Clarification on these points is essential for accurate problem-solving in physics.
My problem is with the solution
If we differentiate:
dU/dx = 9x^2 - 7
From where does the negative come from also the j component?
- #2
voko
- 6,053
- 391
You should have used the homework template, then you have written the relevant equations. What equation is relevant for finding force from potential energy?
- #3
omarMihilmy
- 31
- 0
F = dU/dx
- #4
voko
- 6,053
- 391
That is not correct.
- #5
omarMihilmy
- 31
- 0
Correct me if I am wrong
- #6
shanname
- 8
- 0
Why are you disregarding the y-component when you differentiate?
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question.
Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point?
Lets call the point which connects the string and rod as P.
Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
This problem is two parts. The first is to determine what effects are being provided by each of the elements - 1) the window panes; 2) the asphalt surface. My answer to that is
The second part of the problem is exactly why you get this affect.
And one more spoiler:
Let's declare that for the cylinder,
mass = M = 10 kg
Radius = R = 4 m
For the wall and the floor,
Friction coeff = ##\mu## = 0.5
For the hanging mass,
mass = m = 11 kg
First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on.
Force on the hanging mass
$$mg - T = ma$$
Force(Cylinder) on y
$$N_f + f_w - Mg = 0$$
Force(Cylinder) on x
$$T + f_f - N_w = Ma$$
There's also...
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