Work and power needed to pull a 70 kg skier up a frictionless 30° slope

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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
 
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?
 
All right, seems like you did OK, then:
For b) follow the procedure I indicated..
 
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?
 
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..)
 
nevermind, i got it... yipee! time for a drink : )