Conservation of Energy ball of mass

[KatlynEdwards]

Homework Statement

A ball of mass m falls from height h_i to height h_f near the surface of the Earth. When the ball passes h_f it has speed v_f. Ignore air resistance, and assume that any changes in kinetic energy of the Earth are negligible. Also, the starting velocity of the ball is zero.

Write an expression for each of the following quantities in terms of the given variables and any physical constants. If any of these are zero, state so explicitly.


1. The change in the kinetic energy of the system (the ball and the earth).
2. The change in potential energy of the system.
3. The net external work by external forces on the system
4. Now write an equation that relates the expressions above and use it to solve for the final speed of the ball.

Homework Equations

KE = 1/2 m*v^2
PE = -m*g*h

The Attempt at a Solution

1. So the change in kinetic energy of the system is 1/2 m v^2
2. the change in potential energy is -m*g*h
3. The net external work is zero, because all the changes are internal
4. So I would say 1/2 m*v^2 -m*g*h = 0 which is the total net external work.

Solving for v I get the square root of (2*g*h)...
Although I don't recognize this formula.


Can anyone tell me if I'm doing something wrong?
Thanks!

 

[thrill3rnit3]

Everything looks good.

Don't worry, you'll see $$\sqrt{2gh}$$ a lot when dealing with those kinds of questions.

 

[Redbelly98]

Homework Statement

A ball of mass m falls from height h_i to height h_f near the surface of the Earth. When the ball passes h_f it has speed v_f. Ignore air resistance, and assume that any changes in kinetic energy of the Earth are negligible. Also, the starting velocity of the ball is zero.

Write an expression for each of the following quantities in terms of the given variables and any physical constants. If any of these are zero, state so explicitly.


1. The change in the kinetic energy of the system (the ball and the earth).
2. The change in potential energy of the system.
3. The net external work by external forces on the system
4. Now write an equation that relates the expressions above and use it to solve for the final speed of the ball.

Homework Equations

KE = 1/2 m*v^2
PE = -m*g*h

The Attempt at a Solution

1. So the change in kinetic energy of the system is 1/2 m v^2
2. the change in potential energy is -m*g*h
3. The net external work is zero, because all the changes are internal
4. So I would say 1/2 m*v^2 -m*g*h = 0 which is the total net external work.

Solving for v I get the square root of (2*g*h)...
Although I don't recognize this formula.


Can anyone tell me if I'm doing something wrong?
Thanks!

1 & 2
I disagree with your answers, because they are asking for the change in the kinetic and potential energy. You simply give the formulas for the kinetic and potential energies, which is different.

So...

1. What is the initial kinetic energy? And what is the final kinetic energy? Use that to get the change in kinetic energy.

2. What is the initial potential energy? And what is the final kinetic energy? Use that to get the change in kinetic energy.

3. I agree, the external work is zero.

 

[KatlynEdwards]

Well if they want the change in potential and kinetic energy of the system, I would have to say zero because of the fact that there is no external work.
So would I say:
ΔK + ΔP = External Work
0 + 0 = 0 ?if the system was just the ball, instead of the ball and the earth, my first two answers would be correct, and then the external work would not be equal to zero. Right?

 

[Redbelly98]

Not quite.

ΔK + ΔP = 0, but that does not mean that ΔK=0 and ΔP=0.

Think about it: given that the height changed from h_i to h_f, what is ΔP?

 

[KatlynEdwards]

Well, ΔP = m*g*Δh
= m*g*(h_f - h_i

ΔK = 1/2 m*v^2
= 1/2 m*v_f^2

So basically ΔP and ΔK are opposites?

 

[Redbelly98]

Looks good! And yes, ΔP = -ΔK, when total energy is conserved (no external forces).