Solving Physics Homework: Mass, Slope & Speed | POWER Needed

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A skier with a mass of 70 kg is pulled up a 30º slope for 60 m at a constant speed of 2 m/s, requiring calculations for work and power. The work needed is determined by the potential energy change, using the formula W = mgy, where height is calculated from the slope. There was confusion regarding the angle of the slope and the unit conversion from joules to kilojoules. The discussion also highlighted the importance of correctly applying the work formula and understanding average force. Ultimately, the user resolved their issues and successfully calculated the power needed for the task.
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POWER! its no good

alright, I'm stuck with a homework problem again!


A skier of mass 70kg is pulled up a slope by a motor-drive cable.
A) homework much work is required to pull him 60 m up a 30º slope at a contant speed of 2m/s (frictionless)?
B) what power must a motor have to perform this task?

please help. soon.
 
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maybe more specifically, I'm having a lot of trouble deciding what equation to use for work... someone want to give me a push start?
 
A) Since your kinetic energy doesn't change, you only need to consider the work necessary to increase the potential energy of the system.
B) Find the average force applied from your expression for the work done in A)
 
i guess no one wants to help me tonight... okay, thanks anyway.
 
Were you losing hope, then?..:wink:
 
sorry, i didnt see that post! so, are you talking potential energy then? like W = mgy which is W = (70)(9.8)(30)
I got y, aka h, from 60sin20. i try this and i get the wrong answer. the answer is 21kJ
 
I AM A COMPLETE IDIOT! i have been doing this problem all night and the answer is right. i guess I'm just having calculator input errors
 
anymore help with part b?
 
Three questions:
a) Is the slope 20 degrees or 30 degrees?
(You have indicated both..)

b)Have you remembered that the answer is given in kJ rather than J, which is what you will get from your calculations.
c) Perhaps you should try "up" meaning a vertical displacement of 60 meters?
 
  • #10
All right, seems like you did OK, then:
For b) follow the procedure I indicated..
 
  • #11
i'm getting the wrong answer-- the answer is .92hp... i don't know if I'm converting wrong or what. okay, you suggested average force? I'm sorry, I'm really stupid-- what's the equation for that?
 
  • #12
W=Fd, where d is the traversed distance.
IConvert W into joules perhaps; I've no idea how watts are converted into horsepower..:redface:
(Have only used N*m/s, which is watts I would think..)
 
  • #13
nevermind, i got it... yipee! time for a drink : )
 
  • #14
To your health..
 
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