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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?

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- Thread starter qman316
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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?

- #2

tiny-tim

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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?

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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.

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tiny-tim

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Hi qman316!

But you

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

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how i would calculate it is to multiply the force from the horse by the distance travelled (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?

am i missing something?

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

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tiny-tim

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how i would calculate it is to multiply the force from the horse by the distance travelled (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

oh … and this isn't your thread, so you

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HallsofIvy

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

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tiny-tim

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You left out the angle!

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Kurdt

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If its still wrong after that check the units.

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ur answer should probably be in newton meters

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- 555

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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?

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?

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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.

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Kurdt

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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

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well wouldnt 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?

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?

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tiny-tim

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well wouldnt 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

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so what am i doing wrong? its been a little while since of done problems like this and i dont have equations in front of me im 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 ya blame me for concentrating on the horse too much haha the question asks what is the work done by the horse

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

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

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

The equation for work is:

[tex]W = \vec{F} \cdot \vec{d} = Fd \cos\theta[/tex]

In this equation, F is the total force.

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ok so based on that i was right...

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i got an answer of ******

anybody confirm? or do i have a unit error?

anybody confirm? or do i have a unit error?

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