Power problem -- Mule pulling a wagon, convert work and power to time

In summary, the conversation revolves around calculating the time interval for a mule to pull a load of gold on a wagon, given its power output of 746 watts and the total work done by the team on the wagon being 23,500,000 joules. The formula used is power = work/time, and after rearranging it, the final answer is calculated to be approximately 1.44 seconds. The conversation also highlights the importance of memorizing formulas and understanding the units of measurement.
  • #1
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Homework Statement



a mule is pulling a load of gold on a wagon. If the mules power output is 746 watts. if in a certain time interval the total work done by the team on the wagon was 23,500,000 j, how long was that interval of time

Homework Equations


I think its power = work divided by time

The Attempt at a Solution


I don't have one
 
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  • #2
Kaneki said:

Homework Statement



a mule is pulling a load of gold on a wagon. If the mules power output is 746 watts. if in a certain time interval the total work done by the team on the wagon was 23,500,000 j, how long was that interval of time

Homework Equations


I think its power = work divided by time

The Attempt at a Solution


I don't have one

Why don't you try to use power=work/time? What is a watt in units of joules? Just try.
 
  • #3
Dick said:
Why don't you try to use power=work/time. Just try.
that's the thing, I don't know the numbers. The time isn't mentioned so I would have to re arrange the formula. My guess would be work times power or work divided by time. Like 746 times 746 times 23,500,000 or divided that. I am dumb and I can't figure it out.
 
  • #4
Kaneki said:
that's the thing, I don't know the numbers. The time isn't mentioned so I would have to re arrange the formula. My guess would be work times power or work divided by time. Like 746 times 746 times 23,500,000 or divided that. I am dumb and I can't figure it out.

I'll rearrange it for you. If power=work/time then time=work/power. I'm dumb is pretty weak excuse. You have to do something here. Now try and do it. Look things up if you have to.
 
  • #5
Dick said:
I'll rearrange it for you. If power=work/time then time=work/power. I'm dumb is pretty weak excuse. You have to do something here. Now try and do it. Look things up if you have to.
Would we have have to convert 746 watts into jouls? It would be 3,600,000. So will it be 23,500,000/3,600,000 = 1.44? I am sorry, it's not an excuse but it's hard for me to actually just sit down and think. Thats why I won't do AP physics next year.
 
  • #6
Kaneki said:
Would we have have to convert 746 watts into jouls? It would be 3,600,000. So will it be 23,500,000/3,600,000 = 1.44? I am sorry, it's not an excuse but it's hard for me to actually just sit down and think. Thats why I won't do AP physics next year.

Sorry, but if this one is a challenge then you probably shouldn't try AP physics. A watt is a joule/second. This is actually pretty elementary material. It's just dealing with basic algebra and physics units. I don't know what to say.
 
  • #7
The time isn't mentioned so I would have to re arrange the formula. My guess would be work times power or work divided by time.

The formula you quoted was..

power(W) = work(J) divided by time(S)

So rearranging that gives..

Time(S) = work(J) divided by power(W).
 
  • #8
Kaneki said:
Would we have have to convert 746 watts into jouls? I
Mastering introductory physics is initially made a little more difficult because units of many quantities go by a number of names.

The Joule is another name for a Newton.metre, and you can remind yourself of this by memorizing the formula W = F.s
The Watt is another name for a Joule/second or a Newton.metre/second, this comes from the formula P = W/t

So there is a second reason for memorizing formulae---not only are they needed for calculations, but once you realize how they also help you with the names of units, equations are immensely valuable tools of trade, whatever area of science you are dealing with.

BTW, that 746 conversion factor is for hp to watts, but there is no mention of horsepower anywhere here. You are already given the mule's power in watts.
 
Last edited:

1. How do you calculate power in the context of a mule pulling a wagon?

In this scenario, power is calculated by multiplying the force exerted by the mule on the wagon by the speed at which it is moving.

2. What unit is power typically measured in?

Power is typically measured in watts (W) or horsepower (hp).

3. How can you convert power to time in the context of a mule pulling a wagon?

To convert power to time, you would need to know the distance the wagon needs to travel and the power of the mule. You can then use the equation Time = Work/Power to calculate the time it would take for the mule to pull the wagon a certain distance.

4. Can you explain the relationship between work and power in this scenario?

Work and power are directly related in this scenario. Work is the amount of force needed to move the wagon a certain distance, and power is the rate at which that work is being done. The more power the mule exerts, the faster the wagon will be pulled and the less time it will take.

5. How does the incline or terrain affect the power needed for the mule to pull the wagon?

The incline or terrain can greatly affect the power needed for the mule to pull the wagon. If the wagon is going uphill, the mule will need to exert more power to overcome the force of gravity. Alternatively, if the terrain is rough or uneven, the power needed may also increase as the mule has to navigate through obstacles.

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