Calculating Work and Kinetic Energy: A Ski Slope Scenario

[kappcity06]

work and kinetic energy?

i was wondering if anyone could help with this proplem:

A skier pulled by a tow rope up a frictionless ski slope that makes an angle of 12° with the horizontal. The rope moves parallel to the slope with a constant speed of 1.0 m/s. The force of the rope does 930 J of work on the skier as the skier moves a distance of 7.2 m up the incline.

If the rope moved with a constant speed of 2.0 m/s, how much work would the force of the rope do on the skier as the skier moved a distance of 7.2 m up the incline?

At what rate is the force of the rope doing work on the skier when the rope moves with a speed of 1.0 m/s

At what rate is the force of the rope doing work on the skier when the rope moves with a speed of 2.0 m/s?


all that i have is w=f*d*cos(theta) please help

 

[kappcity06]

ok i tried sumation of work=1/2mv(final)^2-1/2mv(inital)^2. no go. then i tried to used w=fdcos(theata) 930=f*7.2*cos(12). nothing. does speed matter when dealing with force and work?!?

 

[nrqed]

ok i tried sumation of work=1/2mv(final)^2-1/2mv(inital)^2. no go. then i tried to used w=fdcos(theata) 930=f*7.2*cos(12). nothing. does speed matter when dealing with force and work?!?

The formula w=fdcos(theta) is perfectly fine as well as sum of the work = final kinetic energy - initial kinetic energy.

so the work done by the rope is T d cos (theta) where T is the tension in the rope and theta is the angle between the rope and the direction of the tension, right? So the question becomes this: if the skier is pulled at constant speed at 1 m/s or at 2 m/s, what does it imply for the tension in th erope? . Is it the same at 1m/s than it is at 2 m/s? That will andwer your question.

For the power (or rate of work) of a force, you must divide the work done by the time during which the force was in action.

 

[kappcity06]

thank you very much