Finding Work and Force in a Thrown Ball Problem

[Rumplestiltskin]

Homework Statement

A 4.0kg ball is thrown vertically up into the air with an initial velocity of 8.5ms⁻¹ . By the time it is height h metres above the starting point, it has a velocity of 3.0ms⁻¹ and has done 4.0J of work against air resistance. Find h.

Homework Equations

Work done = force x distance

The Attempt at a Solution

Work/force = distance. I have the work done, I need the force.
Force = 4kg * [ (3 - 8.5) / t ]
How am I supposed to find t without the distance?

 

[Doc Al]

Use the work-energy theorem.

 

[TSny]

If you use energy concepts, you will not need to know any forces or the time.

How much mechanical energy does the ball have immediately after it is thrown?

 

[Rumplestiltskin]

If you use energy concepts, you will not need to know any forces or the time.

How much mechanical energy does the ball have immediately after it is thrown?

KE = 0.5 * 4 * 8.5² = 144.5 J
I can't convert this to GPE without assuming that h is the max height.

 

[Chestermiller]

KE = 0.5 * 4 * 8.5² = 144.5 J
I can't convert this to GPE without assuming that h is the max height.

Take as the datum of GPE the release point of the ball.

 

[TSny]

The initial mechanical energy does not get converted entirely to GPE. Note that the ball still has a speed of 3 m/s at the final position. Also, some of the mechanical energy is used to do work against air resistance.

 

[Ray Vickson]

Homework Statement

A 4.0kg ball is thrown vertically up into the air with an initial velocity of 8.5ms⁻¹ . By the time it is height h metres above the starting point, it has a velocity of 3.0ms⁻¹ and has done 4.0J of work against air resistance. Find h.

Homework Equations

Work done = force x distance

The Attempt at a Solution

Work/force = distance. I have the work done, I need the force.
Force = 4kg * [ (3 - 8.5) / t ]
How am I supposed to find t without the distance?

You should realize that your equation "Work done = force x distance" is FALSE if the force varies over the interval you are looking at. So, the equation fails if the force $F$ is either a function of time $t$ or position $x$. In general, you need to do an integral. If the force varies, the work done between times $t_1$ and $t_2$ or positions $x_1 = x(t_1)$ and $x_2 = x(t_2)$ is
$$\text{Work done} = \int_{t_1}^{t_2} F(t) x'(t) \, dt \; \Longleftarrow \: F = F(t)$$
or
$$\text{Work done} = \int_{x_1}^{x_2} F(x) \, dx \; \Longleftarrow \: F = F(x)$$

 

[lychette]

Homework Statement

A 4.0kg ball is thrown vertically up into the air with an initial velocity of 8.5ms⁻¹ . By the time it is height h metres above the starting point, it has a velocity of 3.0ms⁻¹ and has done 4.0J of work against air resistance. Find h.

Homework Equations

Work done = force x distance

The Attempt at a Solution

Work/force = distance. I have the work done, I need the force.
Force = 4kg * [ (3 - 8.5) / t ]
How am I supposed to find t without the distance?

4kg is not a force !

 

[Doc Al]

In general, you need to do an integral.

Luckily, the integration of the air resistance force has already been done.

 

[Ray Vickson]

Luckily, the integration of the air resistance force has already been done.

Right. But the OP wrote Force × distance, as though he believes that is always true. I was trying to get him to realize that is not the case in general, but of course it is SOMETIMES true.

 

[Rumplestiltskin]

4kg is not a force !

Didn't say it was.

Right. But the OP wrote Force × distance, as though he believes that is always true. I was trying to get him to realize that is not the case in general, but of course it is SOMETIMES true.

I'm too novice, too young and too stupid to understand that, but thanks. I'll do the Q in my own time.

 

[Doc Al]

KE = 0.5 * 4 * 8.5² = 144.5 J
I can't convert this to GPE without assuming that h is the max height.

You don't need to 'convert' this to anything. It's the KE at the point of release.

What's the total mechanical energy at the point of release? At the height h? What's the change in total energy?

 

[Rumplestiltskin]

You don't need to 'convert' this to anything. It's the KE at the point of release.

What's the total mechanical energy at the point of release? At the height h? What's the change in total energy?

KE = 0.5 * 4 * 3² = 18 J
18 - 144.5 + 4 = -122.5 J
GPE = -122.5 = 4 * -9.8 * h
...
h = 3.1m, correct answer. Thanks.

 

[TSny]

You need to rethink the signs. In the expression GPE = mgh, g is a positive quantity. The GPE should be positive at the height h.

 

[Rumplestiltskin]

You need to rethink the signs. In the expression GPE = mgh, g is a positive quantity. The GPE should be positive at the height h.

So delta KE isn't negative either, why not?

 

[Doc Al]

So delta KE isn't negative either, why not?

ΔKE is certainly negative since it slows down.

 

[Doc Al]

Set it up carefully, and the signs will work out:

Initial Energy = ?
Final Energy = ?
Δ Energy = Final Energy - Initial Energy = ?

 

[Rumplestiltskin]

Set it up carefully, and the signs will work out:

Initial Energy = ?
Final Energy = ?
Δ Energy = Final Energy - Initial Energy = ?

144.5 J
18 J
18 - 144.5 = -126.5 J
GPE = -126.5 = 4 * 9.8 * h
h must be negative?

 

[TSny]

You are still writing the GPE at height h as a negative number. As an object gains height, it gains GPE. Since you are taking GPE = 0 at the starting point, then GPE must be positive at height h.

The total mechanical energy at any point is E = KE + GPE. If there were no air resistance, the total mechanical energy, E, would not change. However, work had to be done against the air resistance and this uses up some of the mechanical energy (by converting some of the mechanical energy to heat energy, etc.).

You know the value of E at the start. What is the value of E at height h? (Answer this based on the value of E at the start and the amount of work done against air resistance.)

 

[Doc Al]

144.5 J

That's the initial KE. Since we can measure GPE from that starting point, the initial GPE = 0. So yes, that's the initial energy.

18 J

That's the final KE, but what about GPE? (Symbolically, what's the GPE at height h?)

Please read over TSny's posts; he's got you covered.

 

[Rumplestiltskin]

You are still writing the GPE at height h as a negative number. As an object gains height, it gains GPE. Since you are taking GPE = 0 at the starting point, then GPE must be positive at height h.

The total mechanical energy at any point is E = KE + GPE. If there were no air resistance, the total mechanical energy, E, would not change. However, work had to be done against the air resistance and this uses up some of the mechanical energy (by converting some of the mechanical energy to heat energy, etc.).

You know the value of E at the start. What is the value of E at height h? (Answer this based on the value of E at the start and the amount of work done against air resistance.)

Ah, so although change in KE is negative, when you equate this to GPE your value is positive because GPE is a scalar quantity.

KE at h is 18 J.
GPE at h is 122.5 J.
(18 + 126.5) - 4 = 140.5 J.

It's helpful to know the stuff about the signs, but did I need to find the value of E? Found h without it.

 

[Doc Al]

KE at h is 18 J.

OK.

GPE at h is 122.5 J.

Since h is what you are trying to solve for, do not assume a value. Express GPE symbolically.

Eventually you want this equation: E_initial - Work done against air resistance = E_final

 

[Rumplestiltskin]

Since h is what you are trying to solve for, do not assume a value. Express GPE symbolically.

Eventually you want this equation: E_initial - Work done against air resistance = E_final

How did I assume a value when KE = GPE?
4 * 9.8 * h = GPE
39.2h. Not sure if that's what you mean.

 

[Doc Al]

How did I assume a value

You assumed a value when you wrote:

GPE at h is 122.5 J.

I was looking for GPE = mgh.

 

[Rumplestiltskin]

You assumed a value when you wrote

How? What I've been taught is that KE = GPE. I want to know why this is wrong on a technicality.

 

[Doc Al]

How? What I've been taught is that KE = GPE. I want to know why this is wrong on a technicality.

I'm not sure where you are going with this. To solve for "h" you must have an expression with "h" in it at some point. GPE = mgh is that expression.

And I think you meant ΔGPE = -ΔKE, which would be true except when dissipative forces act (like air resistance in this example).

 

[Rumplestiltskin]

I'm not sure where you are going with this. To solve for "h" you must have an expression with "h" in it at some point. GPE = mgh is that expression.

And I think you meant ΔGPE = -ΔKE, which would be true except when dissipative forces act (like air resistance in this example).

You have a problem with me jumping to the calculations instead of stating the equation first. Why is this important?
-ΔKE? So if the object were accelerating upwards and the KE was increasing, the GPE would decrease?
So my value was an assumption because I don't know to what proportion the air resistance acts on either KE or GPE?

 

[Doc Al]

You have a problem with me jumping to the calculations instead of stating the equation first. Why is this important?

In order for others to provide help, you have to show how you arrived at any calculations you did. Just stating a number without the reasoning behind it is meaningless.

-ΔKE? So if the object were accelerating upwards and the KE was increasing, the GPE would decrease?

No. ΔGPE = -ΔKE only applies when mechanical energy is conserved: When no dissipative or other non-conservative forces act. But it is irrelevant here. (I was just correcting your general statement, GPE = KE, which is incorrect.)

So my value was an assumption because I don't know to what proportion the air resistance acts on either KE or GPE?

There is no need for any assumptions.

 

[Rumplestiltskin]

In order for others to provide help, you have to show how you arrived at any calculations you did. Just stating a number without the reasoning behind it is meaningless.

I'm confused -- if you look back, you'll find that I equated the GPE to values for m * g * h. I didn't express that algebraically, but other posters had figured.

There is no need for any assumptions.

I'm not being defensive here, but I still don't understand how the value 122.5 J is an assumption. The change in KE was 18 - 4 - 144.5 = -122.5 J. By your own equation, this means that the GPE is 122.5. So I'm wrong about the change in KE. Am I prematurely taking away the work done against air resistance, which should instead be taken away from the mechanical energy? I'd understand it if you said that "the change in KE is just 18 - 144.5 = -126.5J. This would make the GPE 126.5J not considering air resistance. Considering air resistance, it is unknown how much it detracts from either KE or GPE, so the value 122.5 is 'assumed'. "

 

[Jack Weatherilt]

Using ${1 \over 2} m v^2$, we get the initial energy of the ball to be:
$${1 \over 2} \times 4 \times 8.5^2 = 144.5$$
However, at $h$ the ball still has velocity (8.5 ms^-1), as well as having done work against air resistance (the 4 J mentioned in the question). This leaves the remaining energy at:
$$144.5 - ({1 \over 2} \times 4 \times 3^2 + 4) = 122.5$$
Assuming that the rest of the energy must be GPE, i.e. $mgh$, we can divide 122.5 by $mg = 4 \times 9.81$, giving us the final value for $h$ as 3.125 m.

Since this is an Isaac Physics question, you'll want to be giving you answer to 2 SF, i.e. 3.1 m.