Formulas converting Potential Energy to Kinetic Energy

[magnetics]

Homework Statement

A 5kg ball (m), is to be dropped 1 meter (h) in a vacuum with no resistance in the Earth's gravity (g).

Use the speed and motion equations to show that its KE at impact is equivalent to its PE at its starting point.

Homework Equations

PE = mgh
KE= 1/2mv^2
S = vt + 1/2gt^2
S = vt

The Attempt at a Solution

Since h = 1
PE = mg

To calculate KE, we need to calculate the time taken to fall 1 meter and hence calculate the velocity.

h = 0 + 1/2gt^2
t^2 = 2/g

S = vt
v = h/t
therefore
v = √g/2
and
v^2 = g/2

KE = 1/2mv^2
KE = 1/2 x mg/2
KE = mg/4

I've now done this a few times and by the above calculations, the KE is 1/4 of the PE.

There has to be an error somewhere in my calculations, hopefully someone help enlighten me :)

 

[majormaaz]

S = vt + 1/2gt^2

I've now done this a few times and by the above calculations, the KE is 1/4 of the PE.

There has to be an error somewhere in my calculations, hopefully someone help enlighten me :)

Try using (v_t)² = (v₀)² + 2aΔx. Your 1/4 problem should be gone then

 

[BvU]

Homework Statement

A 5kg ball (m), is to be dropped 1 meter (h) in a vacuum with no resistance in the Earth's gravity (g).

Use the speed and motion equations to show that its KE at impact is equivalent to its PE at its starting point.

Homework Equations

PE = mgh
KE= 1/2mv^2
S = vt + 1/2gt^2
S = vt

I remember something like s = s₀ + v₀t+ 1/2 a t²
And in this case s₀ is 1 m, v₀ = 0 and a = -g

Since h = 1
PE = mg
...
h = 0 + 1/2gt^2
t^2 = 2/g

both OK (if interpreted somewhat liberally)

"S = vt" is NOT ok, though. This is only if v is constant, and v is not constant: there is a constant acceleration (namely -g)

"v = h/t" idem, and the "therefore v = √g/2 and v^2 = g/2" needs to be reviewed.

(What you want here is something like v = v₀ + a t with v₀ = 0 and a = -g )


" KE = 1/2mv^2 " true again, now we only need a better v .

 

[magnetics]

I remember something like s = s₀ + v₀t+ 1/2 a t²
And in this case s₀ is 1 m, v₀ = 0 and a = -g

both OK (if interpreted somewhat liberally)

Thanks BvU.

v = at makes all the difference. You are correct, during the fall, a is constant, not v!

What do you mean by (if interpreted somewhat liberally)?

 

[BvU]

s = s0 + v0t+ 1/2 a t² ; fill in
(in a coordinate system where up is positive, so g = -9.81 m/s²):
0 = h - 1/2 9.81 t² bring to other side:
1/2 9.81 m/s² t² = h

The liberties are swapping s and s0 and the sign of g.
You have the benefit of the doubt (it comes out all right this time, but is that by coincidence?):

• I can see you take g = +9.81 m/s², because you end up with t²=2h/g
  -- also missed a factor h there: h is 1, so the value isn't the issue but you need the dimension of length to get s² --
• but if you take g positive, you should really take h negative, )

but mistakes with signs and dimensions are made very easily, so be careful. Better to work explicitly and conscientiously.

 

[majormaaz]

[*]I can see you take g = +9.81 m/s², because you end up with t²=2h/g
-- also missed a factor h there: h is 1, so the value isn't the issue but you need the dimension of length to get s² --
[*]but if you take g positive, you should really take h negative, ) [/LIST]but mistakes with signs and dimensions are made very easily, so be careful. Better to work explicitly and conscientiously.

Just a note - if you're solving for t², then making h negative if you have gravity as positive is totally fine. However, if you're solving for t, just a hint that you can't take the root of a negative expression and that you can do something about it.