What is the average power of the horse?

[qman316]

I have been working on this problem for awhile now and I can't get anywhere. Can anyone help?

1) A horse pulls a cart with a force of 155.0 N at and angle of 32.0o with respect to the horizontal and moves along at a speed of 13.8 km/hr. How much work does the horse do in 8.6 min?

2) What is the average power of the horse?

 

[tiny-tim]

Welcome to PF!

Hi qman316! Welcome to PF!

(I assume it means that the road is horizontal, but the angle of the force from the horse is 32º)

1) Do you know how to calculate work?

 

[rockerdoctor]

First off, draw a picture. Once you have that think of what work and power really are. how do you find work?
how do you find power?
what equations do you use?

i also think it may be better to figure out number 2 before number one. but that my be just me.

 

[qman316]

All the equations I have to figure out Work involve knowing the distance, which I do not know in this problem. Is there another way to find work without knowing a distance?

 

[leright]

hmmm...well since the horse and cart are moving at a constant speed then the NET work is zero (the work the horse does and the work friction does cancel out resulting in no change in KE and thus no acceleration). However, the work the horse does is found by projecting the line of force caused by the horse onto the direction of motion. You would then multiply this by the speed to get the power (rate at which work is done). You can then multiply the power by the time (you need to convert minutes to hours though) to get work.

 

[rockerdoctor]

not all the equations to find work requires distance. do you know what the formula for kinetic energy is?

 

[tiny-tim]

All the equations I have to figure out Work involve knowing the distance, which I do not know in this problem. Is there another way to find work without knowing a distance?

Hi qman316!

But you do know the distance …

you know the time and the speed, so the distance is … ?

 

[shamrock5585]

how i would calculate it is to multiply the force from the horse by the distance traveled (which is just the velocity multiplied by the time)... F*d=Work

am i missing something?

the angle really has nothing to do with it unless your figuring out the work done by gravity right?

 

[tiny-tim]

Hi shamrock5585!

how i would calculate it is to multiply the force from the horse by the distance traveled (which is just the velocity multiplied by the time)... F*d=Work

am i missing something?

Yes … you're missing how horses and carts work!

(how old is this book? )

oh … and this isn't your thread, so you won't give the answer, will you?

 

[qman316]

so... d=v*t which is 13.8km/hr(.1433hr), d=1.978 then F*d=W which is 155N*1.978=306.59, what did I do wrong because it says 306.59 is the wrong answer?

 

[HallsofIvy]

The horse moves at a speed of 13.8 km/hr. Since your are asked for average power, not work, you can start by assuming any time you like- say 1 hour. How far does the horse pull the cart in one hour?

 

[qman316]

the question in part a asks for the work and b the average power

 

[tiny-tim]

Hi qman316!

so... d=v*t which is 13.8km/hr(.1433hr), d=1.978 then F*d=W which is 155N*1.978=306.59, what did I do wrong because it says 306.59 is the wrong answer?

You left out the angle!

 

[Kurdt]

If its still wrong after that check the units.

 

[shamrock5585]

ur answer should probably be in Newton meters

 

[shamrock5585]

i was suggesting that the angle did not mean anything because i assumed the force was at an angle of 32 degrees instead of the force being horizontal and the box moving up a hill at angle 32 degrees...

 

[Nick89]

Either if the horse is walking on a slope of 32 degrees, or if the 'connection' from the horse to the cart is at an angle of 32 degrees, you still need to use it!I assume it is meant that the connection is not mounted horizontally between the horse and the cart. Then, the force that is given is the total force, at an angle of 32 degrees.

What is the equation for work? I saw you use "W = Fd" before but that is wrong! Can you see from this what you did wrong?

 

[Nick89]

What is the equation for work? I saw you use "W = Fd" before but that is wrong! Can you see from this what you did wrong?

Let me just correct myself there (since I cannot seem to edit my post anymore).
The equation 'W = Fd' is valid, but only under a certain condition, which has to do with the direction of the force.

 

[Kurdt]

Let me just correct myself there (since I cannot seem to edit my post anymore).

Editing time is now 30 minutes. See: https://www.physicsforums.com/showthread.php?t=228058

 

[shamrock5585]

well wouldn't W=Fd still be valid but force would be Fcos32 now?

to clarify what i meant before is that if the horse was walknig up the path the angle does not matter... the angle between force and displacement is 0 so Fcos0=F because cos0 =1

so the horse is walking horizontal with a rope tied to a box pulling it up a slope of angle 32 degrees?

 

[tiny-tim]

well wouldn't W=Fd still be valid but force would be Fcos32 now?

to clarify what i meant before is that if the horse was walknig up the path the angle does not matter... the angle between force and displacement is 0 so Fcos0=F because cos0 =1

so the horse is walking horizontal with a rope tied to a box pulling it up a slope of angle 32 degrees?

Hi shamrock5585!

You're concentrating too much on the horse.

The horse is connected to the cart by a shaft or rope, and the force on the cart (and the reaction force on the horse) comes entirely from the tension in the shaft or rope, which will be along its length, at an angle of 32º.

 

[shamrock5585]

so what am i doing wrong? its been a little while since of done problems like this and i don't have equations in front of me I am just going from my head... i was pretty sure it was the the force times the cos of the angle between force and dislacement times distance to get work... no?

oh yeah... and can you blame me for concentrating on the horse too much haha the question asks what is the work done by the horse

 

[Nick89]

Here, I quickly drew up an image (don't mind the horse lol) of what I believe is the case:

The 'rope' is attached to the horse under an angle of 32 degrees.

Only the horizontal component $\vec{F}_{hor}$ contributes to actually pulling the cart, so only this component does work.

The equation for work is:
$$W = \vec{F} \cdot \vec{d} = Fd \cos\theta$$

In this equation, F is the total force.

 

[shamrock5585]

ok so based on that i was right...

 

[shamrock5585]

i got an answer of ******

anybody confirm? or do i have a unit error?

 

[tiny-tim]

i got an answer of ****** N*m

anybody confirm? or do i have a unit error?

Hi shamrock5585!

good try … but it's still clearly legible, unfortunately … so I think we'd better neither confirm nor correct your answer until the OP has had a go.

 

[shamrock5585]

haha i tried to put it in better grey but when i went back to edit i couldn't edit the color... as long as the OP knows how to arrive at the number then the answer shouldn't really matter...

qman, did you get something close to this? (make sure you are in Newton meters)

 

[Kurdt]

If you check the forum rules shamrock you will find that giving out answers is clearly forbidden. The answer does really matter since the OP could just use that instead of putting in any work themselves, in which case they learn nothing, and it's even worse if that answer is wrong.

 

[Nick89]

Please don't blatantly post the answer... We are trying to help the OP come up with the answer himself, not simply give it to him!

EDIT
Shamrock this wasn't aimed at you, someone else posted the complete answer but it is now removed :)

 

[shamrock5585]

instead of everybody complaining can we ask... qman do you understand how this is obtained?

 

[Moonbear]

instead of everybody complaining can we ask... qman do you understand how this is obtained?

And if s/he answers "yes?" Let the OP work through the problem themselves with hints and tips from the Homework Helpers, not outright answers. The purpose of this forum is not to make sure everyone has the right answers on their homeworks, but to make sure they've learned HOW they arrived at that answer and fully understand the subject so they can solve problems on their own. Handing an answer out doesn't accomplish that, neither does asking someone a yes/no question if they understand (some students do not realize they don't fully understand if they have an answer and can work backward since they haven't had a chance to work without that crutch). Let them SHOW you they understand by putting their own work into the problem.

Since there have now been TWO attempts from different people to give out an answer/complete solution here, this is an official notice from the staff letting folks know that we do NOT permit giving out complete answers.

Sorry to qman that your thread has been so disrupted.

 

[WillySaw]

why is my earlier post deleted?

 

[Kurdt]

why is my earlier post deleted?

If you didn't get a PM about that, look at Moonbear's post above or review the forum rules which can be found in the link below, which you agreed to when signing up.

https://www.physicsforums.com/showthread.php?t=5374

Pay particular attention to the section labelled homework help.

 

[Nick89]

why is my earlier post deleted?

Because you gave the answer to the problem. The guidelines state that we must try to help the OP by giving him hints and letting him solve the problem himself. Simply giving him the answer is not going to help and is not appreciated.


Now can we go back to helping the OP?

Do you understand what has been said about using only the horizontal component of the given force, because that is the only force doing work?

 

[Moonbear]

Good grief, folks, I've never seen such a mess in a homework thread before! If you can't help without giving out an answer, or you have something to say that is not directly related to helping the OP, STAY OUT OF THIS THREAD! Last warning! I'm getting an itchy trigger finger and am tempted to start giving out temporary bans for anyone who further disrupts the learning process here.