webren
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Hello,
I solved the first part of this problem, but I am a little stuck with finding the second part, which deals with power.
"A skier of mass 70.0 kg is pulled up a slope by a motor-driven cable. (a) How much work is required to pull him a distance of 60.0 m up a 30 degree slope (assumed frictionless) at a constant speed of 2.00 m/s? (b) A motor of what power is required to perform the task?"
In solving for part (a), I drew a free-body diagram and figured out what my x and y force components were:
Fx = -mgsin(30) = 0 (constant speed, so no acceleration)
Fy = n - mgcos(3) = 0
therefore:
Fx = mgsin(30) = (70)(9.80)(sin(30)) = 343 N.
Because Work = F(displacement)(cos(theta)), we can issue that work is = (343)(60) which equals 20580 J or 20.6 KJ.
I understand that Power = work/time, but because there is no time given in this problem, how do we find part (b)?
Any suggestions would be appreciated. Thank you.
I solved the first part of this problem, but I am a little stuck with finding the second part, which deals with power.
"A skier of mass 70.0 kg is pulled up a slope by a motor-driven cable. (a) How much work is required to pull him a distance of 60.0 m up a 30 degree slope (assumed frictionless) at a constant speed of 2.00 m/s? (b) A motor of what power is required to perform the task?"
In solving for part (a), I drew a free-body diagram and figured out what my x and y force components were:
Fx = -mgsin(30) = 0 (constant speed, so no acceleration)
Fy = n - mgcos(3) = 0
therefore:
Fx = mgsin(30) = (70)(9.80)(sin(30)) = 343 N.
Because Work = F(displacement)(cos(theta)), we can issue that work is = (343)(60) which equals 20580 J or 20.6 KJ.
I understand that Power = work/time, but because there is no time given in this problem, how do we find part (b)?
Any suggestions would be appreciated. Thank you.