# Conservation of Energy Problem (Power)

• EnricoHendro
In summary, the cart travels 13.2 meters in 12 seconds for an average velocity equal to half of the final velocity.
EnricoHendro
Homework Statement
A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine shaft by a cable connected to a winch. The shaft is inclined at 30.08 above the horizontal. The car accelerates uniformly to a speed of 2.20 m/s in 12.0 s and then continues at constant speed. (a) What power must the winch motor provide when the car is mov- ing at constant speed? (b) What maximum power must the winch motor provide? (c) What total energy has transferred out of the motor by work by the time the car moves off the end of the track, which is of length 1 250 m?
Relevant Equations
E=K+Ug
P=F.V
y=V.t
W=P.t
Hello there,
I was trying to solve this problem. I have no problem with part A and C. But in part B, my guidebook arrived with different answer. Can anybody point out what my mistake is? I am using the same method as the elevator motor problem which states :
"A 650-kg elevator starts from rest. It moves upward for 3.00 s with constant acceleration until it reaches its cruising speed of 1.75 m/s. (a) What is the average power of the elevator motor during this time interval?"
My second question is, why can't I use the same method with this problem? I mean it basically has the same "situation" right?

Your answer to a) is about ##10kW## and your answer to b) is about ##15kW##? But, for b) the only difference is a small acceleration (much smaller than the gravity down the slope). The answers to a) and b) should be not that much different.

PeroK said:
Your answer to a) is about ##10kW## and your answer to b) is about ##15kW##? But, for b) the only difference is a small acceleration (much smaller than the gravity down the slope). The answers to a) and b) should be not that much different.
yes, because in the previous problem, (the elevator problem that I mentioned) I tried to solve it by first finding its acceleration and then calculating the power, but I got the answer wrong. So I figured, this is more or less a similar kind of problem, so I solve it using the same approach (as the right answer to the elevator problem), that's where I get the 15kWh. What I am confused is when to use the method that I used, and when to try different approach (like figuring the constant acceleration first)?

PeroK said:
Your answer to a) is about ##10kW## and your answer to b) is about ##15kW##? But, for b) the only difference is a small acceleration (much smaller than the gravity down the slope). The answers to a) and b) should be not that much different.
or is it correct if the acceleration is much smaller than the gravity down the slope, I should just use the acceleration approach? (like finding the acceleration first, and then plug in the P=F.V ?

EnricoHendro said:
yes, because in the previous problem, (the elevator problem that I mentioned) I tried to solve it by first finding its acceleration and then calculating the power, but I got the answer wrong. So I figured, this is more or less a similar kind of problem, so I solve it using the same approach (as the right answer to the elevator problem), that's where I get the 15kWh. What I am confused is when to use the method that I used, and when to try different approach (like figuring the constant acceleration first)?
I don't understand what different methods you are talking about. You have a force needed to do work against gravity and/or accelerate an object. Work done is related to force and distance; power is rate of work per unit time.

EnricoHendro
EnricoHendro said:
or is it correct if the acceleration is much smaller than the gravity down the slope, I should just use the acceleration approach? (like finding the acceleration first, and then plug in the P=F.V ?
Your answers to a) and c) look very good: nice and short. The answer to b) should be nice and simple too!

EnricoHendro
chicken scratchings said:
b)
For the first 12 seconds the car has traveled ##\Delta y = \vec{v_\text{avg}} \Delta t##

Since a = constant, the [garbled] is ##\vec{v_\text{avg}} = \frac{v_o + v_f}{2}##

##\delta y = \frac{0 + 2.2}{12}##
##\delta y = 13.2m##
One notes that you have not yet laid out a coordinate system. But it seems that you want the "y" axis to run diagonally along the length of the track. So "y" is the along-the-track distance the car has moved in the first 12 seconds. So far, so good, I am in agreement that the cart travels 13.2 meters in 12 seconds for an average velocity equal to half of the final velocity.
##v_f## is not equal to ##v_0## so we have ##\delta K## non-zero.
Here, you are using ##K## to denote kinetic energy

I tend to be a nit picker when it comes to declaring your variables before use -- computer programmer by trade
$$W_\text{winch} + W_\text{g} \Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_0^2$$$$W_\text{winch} + mgd \cos{90 + \theta} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_0^2$$
Next, you start plugging in numbers before having solved the equation. That is poor strategy, though it will work. Solving the equation symbolically first is easier, less prone to error and may allow you to cancel common factors
$$W_\text{winch} + (950)(9.8)(13.2)]cos{120} = \frac{1}{2}(950)(2.2)^2 - 0$$
Here I am not pleased with the cosine of 120 degrees. The angle is 30.08 degrees, not 30 degrees. You do not round off for significant figures until you are done with your calculations and are reporting a finished result.

Personally, I would have used the sine of 30.08 degrees and inverted the sign on the term. But I see what you are doing. You are taking your angle with reference to a base zero angle pointing directly down. 30 degrees above the horizontal is 120 degrees above the zero angle and the work due by gravity will pick up a negative sign for such an angle due to the cosine going negative.

Second I would have expressed it as the sine of 30.08 degrees and negated the sign on the term.
##W_\text{winch} = 2299 + 61446##
##W_\text{winch} = 63745##
##P=\frac{W_\text{winch}}{\Delta t} = \frac{63745}{12} = 5312## Watt
So you have total energy expended during the 12 seconds and you divide by the 12 seconds to get average power. That looks good.

But what comes next does not follow. You add steady state power after the first 12 seconds to average power during the first 12 seconds to get maximum power at the end of the first 12 seconds.

What is the motivation for this computation:
Max P = 5312 + 10241

It is extremely tedious to transcribe stuff this way. We should not be forced to do it. That's your job.

EnricoHendro
Here's a quick answer to part c):

The work done is KE + GPE gained. KE is ##\frac 1 2 m v^2## and GPE is ##mgh = mgl\sin \theta##. The KE term is negligible compared to the GPE term. With ##\sin 30 = \frac 1 2##, we have
$$W \approx \frac 1 2 mgl = 5.82MJ$$

EnricoHendro
jbriggs444 said:

One notes that you have not yet laid out a coordinate system. But it seems that you want the "y" axis to run diagonally along the length of the track. So "y" is the along-the-track distance the car has moved in the first 12 seconds. So far, so good, I am in agreement that the cart travels 13.2 meters in 12 seconds for an average velocity equal to half of the final velocity.

Here, you are using ##K## to denote kinetic energy

I tend to be a nit picker when it comes to declaring your variables before use -- computer programmer by trade

Next, you start plugging in numbers before having solved the equation. That is poor strategy, though it will work. Solving the equation symbolically first is easier, less prone to error and may allow you to cancel common factors

Here I am not pleased with the cosine of 120 degrees. The angle is 30.08 degrees, not 30 degrees. You do not round off for significant figures until you are done with your calculations and are reporting a finished result.

Personally, I would have used the sine of 30.08 degrees and inverted the sign on the term. But I see what you are doing. You are taking your angle with reference to a base zero angle pointing directly down. 30 degrees above the horizontal is 120 degrees above the zero angle and the work due by gravity will pick up a negative sign for such an angle due to the cosine going negative.

Second I would have expressed it as the sine of 30.08 degrees and negated the sign on the term.

So you have total energy expended during the 12 seconds and you divide by the 12 seconds to get average power. That looks good.

But what comes next does not follow. You add steady state power after the first 12 seconds to average power during the first 12 seconds to get maximum power at the end of the first 12 seconds.

What is the motivation for this computation:It is extremely tedious to transcribe stuff this way. We should not be forced to do it. That's your job.
I see, so the final part is what I did wrong. For the angle, I missread it, I missed the .08, my bad. For the final computation, I add those 2 powers, because I was imagining that the winch is exerting power on the cart for about 5312 watt for the first 12 seconds to get it to accelerate to the final speed, and then it exerts another 10241 watt for the rest of the motion. So I thought the max power should be the sum of those 2. But I was not really sure.

Last edited:
EnricoHendro said:
I see, so the final part is what I did wrong. For the angle, I missread it, I missed the .08, my bad. For the final computation, I add those 2 powers, because I was imagining that the winch is exerting power on the cart for about 5312 watt for the first 12 seconds to get it to accelerate to the final speed, and then it exerts another 10241 watt for the rest of the motion. So I thought the max power should be the sum of those 2. But I was not really sure.
Your 63745J is the work done in the first 12s (I make it a bit more). So when you divided by 12s you got the average power for the first 12s.
By what logic would you add to that the power at constant max speed?

The acceleration in those 12s is constant. What relationship would you expect between the average power over that time and the peak power?

Are you sure it is 30.08 degrees? Are you perhaps misreading a badly written/typed o as 8?

haruspex said:
Your 63745J is the work done in the first 12s (I make it a bit more). So when you divided by 12s you got the average power for the first 12s.
By what logic would you add to that the power at constant max speed?

The acceleration in those 12s is constant. What relationship would you expect between the average power over that time and the peak power?

Are you sure it is 30.08 degrees? Are you perhaps misreading a badly written/typed o as 8?
I see, so the max power is not the average power of the first 12 sec plus the power at constant max speed. Wait a minute, so the max power is the power at constant max speed plus the force times final velocity (which is the "extra power" needed to just accelerate the block at the first 12 sec, and not the full average power for the first 12 sec?) which is approximately 10.6 kW. Am I getting this right?
P.S. I have rechecked, I mistyped, it should be 30.0 only, not 30.08, and by 10.6 kW, I mean the whole max power, not the "extra power" for the first 12 sec.

EnricoHendro said:
plus the force times final velocity
Not sure which force you mean. If it's the extra force to provide the acceleration then yes.
EnricoHendro said:
approximately 10.6 kW
Yes.
Note that it is double the average power over the 12 sec, just as the final speed after constant acceleration from rest is double the average speed.

EnricoHendro said:
I see, so the max power is not the average power of the first 12 sec plus the power at constant max speed. Wait a minute, so the max power is the power at constant max speed plus the force times final velocity (which is the "extra power" needed to just accelerate the block at the first 12 sec, and not the full average power for the first 12 sec?) which is approximately 10.6 kW. Am I getting this right?
P.S. I have rechecked, I mistyped, it should be 30.0 only, not 30.08, and by 10.6 kW, I mean the whole max power, not the "extra power" for the first 12 sec.
This is one of the problems with the plug-and-chug method you've no doubt been taught. As soon as you plug the numbers in you lose sight of the physics. Take a look at the equation: ##P = Fv##. If force is constant, then power is proportional to velocity (for simple accelerated motion at least). That means that:
$$P_{avg} = Fv_{avg} \ \ \text{and} \ \ P_{max} = Fv_{max}$$
Which is a useful observation, and only comes if you resist the urge to plug in the numbers at the first opportunity.

jbriggs444 and EnricoHendro
haruspex said:
Not sure which force you mean. If it's the extra force to provide the acceleration then yes.

Yes.
Note that it is double the average power over the 12 sec, just as the final speed after constant acceleration from rest is double the average speed.
yes, I mean the extra force to provide the acceleration. Thanks a lot man, now I understand

PeroK said:
This is one of the problems with the plug-and-chug method you've no doubt been taught. As soon as you plug the numbers in you lose sight of the physics. Take a look at the equation: ##P = Fv##. If force is constant, then power is proportional to velocity (for simple accelerated motion at least). That means that:
$$P_{avg} = Fv_{avg} \ \ \text{and} \ \ P_{max} = Fv_{max}$$
Which is a useful observation, and only comes if you resist the urge to plug in the numbers at the first opportunity.
ah yes, it is a bad habit of mine. Back when I was in college I did that a lot (I am a Management major though) I study physics by myself, so nobody taught me

## 1. What is the conservation of energy problem?

The conservation of energy problem refers to the principle that energy cannot be created or destroyed, but can only be transformed from one form to another. This means that the total amount of energy in a closed system remains constant over time.

## 2. How is power related to the conservation of energy?

Power is the rate at which energy is transferred or transformed. In the context of the conservation of energy problem, power is important because it determines how quickly energy is being used or changed within a system.

## 3. Can power be created or destroyed?

No, power cannot be created or destroyed. It is a measure of how quickly energy is being transformed, but the total amount of power in a system remains constant.

## 4. How does the conservation of energy problem impact our daily lives?

The conservation of energy problem has a significant impact on our daily lives as it governs the way energy is used and transformed in various systems, such as in our homes, transportation, and industries. It also highlights the importance of using energy efficiently and finding renewable sources of energy to sustain our planet's resources.

## 5. What are some examples of the conservation of energy problem in action?

Some examples of the conservation of energy problem in action include a swinging pendulum, a roller coaster, and a simple light bulb. In all of these scenarios, energy is constantly being transformed from one form to another, but the total amount of energy remains the same.

• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
712
• Introductory Physics Homework Help
Replies
9
Views
326
• Introductory Physics Homework Help
Replies
20
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
471
• Introductory Physics Homework Help
Replies
31
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
607
• Introductory Physics Homework Help
Replies
3
Views
737
• Introductory Physics Homework Help
Replies
11
Views
1K
• Introductory Physics Homework Help
Replies
22
Views
671