- #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?
Neglect gravity other than that of water; neglect friction.
F1=ρgsh=1000*10*1*(1+4)=50000 N;
F1=ρgsh=1000*10*0.1*4=4000 N;
F3=50000-4000=46000 N?
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system
$$M(t) = M_{C} + m(t)$$
$$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$
$$P_i = Mv + u \, dm$$
$$P_f = (M + dm)(v + dv)$$
$$\Delta P = M \, dv + (v - u) \, dm$$
$$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$
$$F = u \frac{dm}{dt} = \rho A u^2$$
from conservation of momentum , the cannon recoils with the same force which it applies.
$$\quad \frac{dm}{dt}...
I was told forces are vectors so they can be divided into x and y components, but what about the normal force or the force of static friction? do you divide those into x and y components if say you had a block that was lying on an angled surface?
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