Conservative/ Non conservative forces problem

AI Thread Summary
The discussion revolves around solving a physics problem involving a rock falling in water, where both gravitational and nonconservative forces are at play. The user is attempting to calculate the nonconservative work done by water resistance, gravitational potential energy, kinetic energy, and total mechanical energy as the rock descends. There is confusion regarding the correct application of formulas, particularly in determining the gravitational potential energy and total mechanical energy at different depths. Clarification is provided that the total mechanical energy should be calculated consistently at the same height, emphasizing the impact of water resistance on energy calculations. The conversation highlights the importance of understanding the problem's requirements and maintaining consistency in energy calculations.
shaggyace
Messages
11
Reaction score
0
Im kind of struggling with some conservative/nonconservative force problems. Someone please help me.

Homework Statement



A 1.8 kg rock is released from rest at the surface of a pond 1.8 m deep. As the rock falls, a constant upward force of 4.3 N is exerted on it by water resistance. Let y=0 be at the bottom of the pond.
Calculate the nonconservative work, W nc, done by water resistance on the rock, the gravitational potential energy of the system, U, the kinetic energy of the rock, K , and the total mechanical energy of the system, E , when the depth of the rock below the water's surface is 0 .

Homework Equations



W=Fd
K=0.5(mv)^2
U=mgd
E=U+K
Wnc=ΔE=Ef-Ei

The Attempt at a Solution



I've been going at this one for almost an hour now. I tried finding the final velocity of the rock first since it starts from rest so that I can find its kinetic energy. To find its non conservative work done by water resistance, I used the Work =force *distance formula and multiplied the force of the water resistance by 1.8 m, but that doesn't seem right. To find its gravitational potential energy, I used the U=mgd formula but got the wrong answer for some reason. For the kinetic energy, I used K=0.5(mv)^2 and used the velocity from the first calculation. To find the total mechanical energy, I added up the potential and kinetic energies. Did I do something wrong? Someone please help me.
 
Physics news on Phys.org
shaggyace said:
Im kind of struggling with some conservative/nonconservative force problems. Someone please help me.

Homework Statement



A 1.8 kg rock is released from rest at the surface of a pond 1.8 m deep. As the rock falls, a constant upward force of 4.3 N is exerted on it by water resistance. Let y=0 be at the bottom of the pond.
Calculate the nonconservative work, W nc, done by water resistance on the rock, the gravitational potential energy of the system, U, the kinetic energy of the rock, K , and the total mechanical energy of the system, E , when the depth of the rock below the water's surface is 0 .

Homework Equations



W=Fd
K=0.5(mv)^2
U=mgd
E=U+K
Wnc=ΔE=Ef-Ei

The Attempt at a Solution



I've been going at this one for almost an hour now. I tried finding the final velocity of the rock first since it starts from rest so that I can find its kinetic energy.
Starting with the final velocity of the rock is not the easiest way to solve this problem. That said, there's nothing keeping you from starting that way, and it may even be useful later to double check your work. What answer did you get? Please show your work of how you got your answer.
shaggyace said:
To find its non conservative work done by water resistance, I used the Work =force *distance formula and multiplied the force of the water resistance by 1.8 m, but that doesn't seem right.
Why not?
shaggyace said:
To find its gravitational potential energy, I used the U=mgd formula but got the wrong answer for some reason.
What answer did you get? Please show your work.

I'm not quite sure how to interpret the problem statement on this particular part. Do you know, are you supposed to show the gravitational potential energy when the rock is at the top or bottom?
shaggyace said:
For the kinetic energy, I used K=0.5(mv)^2 and used the velocity from the first calculation. To find the total mechanical energy, I added up the potential and kinetic energies. Did I do something wrong? Someone please help me.
That last part of the problem statement says, "the total mechanical energy of the system, E , when the depth of the rock below the water's surface is 0." That's different than the total mechanical energy when the rock hits the bottom, because some of the energy was lost to friction by that point (and by the way, when you calculate the rock's total mechanical energy, you need to choose a height and stick with it -- you can't add the mechanical energy when the rock is at the bottom and the gravitational potential energy when the rock is at the top. (If there were no friction, the total mechanical energy would be the same regardless of height, as long as you are consistent with the height. The total mechanical energy is different with different heights, in the case of friction.) In this particular problem, the problem statement tells you to choose the top).
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
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}...
Back
Top