Calculating required horsepower

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Homework Statement:

Calculate the horsepower required to move a 2000 lb. car half a mile in 6 seconds.

Relevant Equations:

1 hp = 550 ft lb / sec

work = distance (ft) * force (lbs)
hp = (distance(ft) * weight(lbs)) / 550(ft lbs/sec)
power = work / time
My attempt at the solution goes as follows:

power = work / time
power = (distance * force) / time
power = (2640 ft * 2000 lb) / 6 seconds = 880000 ft lb / sec

horsepower = 880000 ft lb/sec / 550 ft lb/sec = 1600 hp

however i do not know if this is correct
 

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  • #4
Bystander
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no education in physics or math
Homework Statement: Calculate the horsepower required to move a 2000 lb. car half a mile in 6 seconds.
Homework Equations: 1 hp = 550 ft lb / sec

work = distance (ft) * force (lbs)
hp = (distance(ft) * weight(lbs)) / 550(ft lbs/sec)
power = work / time

horsepower = 880000 ft lb/sec / 550 ft lb/sec = 1600 hp
"Pound-mass" does not equal "pound-force;" can you identify the difference between the two units from the discussion in the article?
 
  • #5
Delta2
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I think the statement uses ##lb## but it actually means ##lbf## since the units for ##1hp## are also given in terms of ##lb##.

Other than that if find the statement a bit vague... What is the force we are pushing against? Is it the car's weight or something else? Also is the speed during the push supposed to be constant?
 
  • #6
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Also is the speed [insert]acceleration(?)[/insert] during the push supposed to be constant?
I'm treating it as a double drag race (two quarters equal a half) at constant acceleration.
 
  • #7
jbriggs444
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power = (2640 ft * 2000 lb) / 6 seconds = 880000 ft lb / sec
That formula computes the energy required to accelerate the car at one gee over a distance of 2640 feet. It then divides by the six second elapsed time to get average power.

One problem is that we have no idea whether one gee is sufficient to move the car 2640 feet in 6 seconds. But we can check for that...

Suppose that the car accelerates at one gee (32 feet per second per second) for a duration of six seconds. Its speed at the end of those six seconds will be 32 * 6 = 192 feet per second. Its average speed will be half that, 96 feet per second. Multiply average speed times duration to get 96 * 6 = 576 feet. Far short of the required half mile.

You will need more than one gee of acceleration to finish the race in six seconds.

Given your level of education, we may not want to tackle a completely correct approach. We should probably assume constant acceleration and work toward determining what final speed is required as a first step.
 
  • #8
DEvens
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Constant acceleration is not constant power. It's actually a much more interesting question if you assume constant power. Kinetic energy is ##E = \frac{1}{2} m V^2## so the power is ## \frac{dE}{dt} = m V \frac{dV}{dt} = m V a = K## a constant. That equation might be solvable by a high-school student that has had intro calculus.
 
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Other than that I find the statement a bit vague...
Vague problem statements are often due to the original Poster paraphrasing the problem, and leaving out important bits without realizing. It would be best for the OP to provide the actual question/problem, word-for-word. Then we can help by showing what the important parts mean.

The OP's proposed solution is clearly heading towards the power required to hoist the car to an elevation of 1/2 mile in the given time; it is highly unlikely that this is what the problem was really about, since that really has nothing to do with a "car."
 
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