Solve Elevator Problem: Work & Power Required

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Homework Help Overview

The problem involves calculating the work and power required to pull a skier up a slope using a motor-driven cable. The skier has a mass of 69.7 kg, and the slope is at an angle of 29.9° over a distance of 59.8 m, with the motion assumed to be frictionless.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the calculation of work using the formula involving force, distance, and angle, with some questioning the correct application of trigonometric functions and the inclusion of mass in the calculations.
  • There are attempts to clarify the relationship between force, mass, and acceleration, with various interpretations of the parameters involved in the problem.
  • Some participants express uncertainty about the correctness of their calculations and seek confirmation on their reasoning.

Discussion Status

The discussion is ongoing, with participants exploring different methods to approach the problem. Some have provided guidance on the importance of including mass in the calculations, while others are still grappling with the correct application of formulas and units. There is no explicit consensus yet, as multiple interpretations and calculations are being examined.

Contextual Notes

There is a noted confusion regarding the correct use of trigonometric functions and the role of mass in determining force. Participants are also addressing unit conversions between watts and horsepower, indicating a need for clarity on these aspects.

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Homework Statement



A skier of mass 69.7 kg is pulled up a slope by a motor-driven cable.
(a) How much work is required to pull him a distance of 59.8 m up a 29.9° slope (assumed frictionless) at a constant speed of 1.90 m/s?
(b) A motor of what power is required to perform this task?
hp

Homework Equations



(a) w=f*change in distance*cos angle
(b) power=work/change in time

The Attempt at a Solution



gravity is the only force in play. so the equation for work would be
w=(9.8m/s^2)(59.8m)(cos29.9)=508.03

take that value and divide it by change in time to get power

*correct?
 
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"cos angle" refers to what angle? With respect to your given angle cos is the wrong trig function. You are also missing a mass factor which would have been obvious if you had checked the units on your answer.
 
No. w=f.s=m.a.s
You haven't taken the mass of the skier into account. The acceleration will be along the incline (which youve taken incorrectly).
 
Power can be calculated by p=fv
 
Ok so the F in the problem is 9.8m/s^2. I would use cos0. However, the change in distance is unknown. Using the formula d=time * speed, I found it to be 8.75m. So, now would I use the work formula with the numbers (9.8m/s^2)(8.75m)(cos0). Which equals 85.57J. After I find that I would divide work by time (5secs) to find average power which is 17.15 W...just my thinking??
 
F is not 9.8 m/s^2. That is a. You are forgetting to take into account mass of the skier. force = mass * acceleration
 
Ok-so hopefully this is it:
W=(mass of skier*acceleration)(change in distance)(cos29.9)
W=(132.43)(59.8)(cos29.9)=6865.23.
Then take that value and divide by the time which is 31.55 seconds.
Which equals 217.6 hp
 
The SI unit of power is watts. There are about 746 watts per h.p. but you better look up that number, i am pulling it out of my head.
 
Is the way I worked the problem out correct? I just need to convert the 217.6 watts to hp? I have had trouble solving this problem. It would be nice to know if this thinking is correct.
 
  • #10
The reasoning seems okay to me.
 
  • #11
UGH! I'm still stuck. Any help would be greatly appreciated!
 
  • #12
Where are you stuck?
 
  • #13
alright...i think I've got it. I have to run to class, but I will post later witht the answer I finally came up with! YAY!
 
  • #14
actually, nevermind that answer was also incorrect. sorry to get our hopes up!:frown: Still working...
 
  • #15
whoops, you shouldn't be looking at the cosine of theta. Sorry, just caught that.
 

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