The answer was 24000 J. So Ritwik06 your calculations were correct because 23600 with sig figs goes to 24000. I still have a question about why you did cos 0 and where you got the zero from. If anyone can answer that for me(where you get the 0 for cos 0) i would appreciate it.
Thank You
~Patrick
Okay, the tension in the cable taking the skier up is m*g*sin0. We never were taught that formula. The displacement along the incline is the 60 m. which is from where skier say gets on chair lift and ends where he gets off chair lift. I've used the W = F * d before but it has always had the cos0...
Homework Statement
A .250 kg block on a vertical spring with a spring constant of 5.00 x 10(3) N/m is pushed downward, compressing the spring .100 m. When released, the block leaves the spring and travels upward vertically. How high does it rise above above the point of release?
Mass = .250...
So the acceleration and force are zero. Then would I do W = (F)(D)(Cos0), W = (60.0)(.82) = 49 J, so my answer would be 49 J? I also could not see the diagram because the site that it is on (imageshack) is blocked at my school.
Thank You
~Patrick
Thanks for the welcome.
I drew a free body diagram and only acting force that I think there is would be Gravitational Force. The only way I know to find gravitational force is the way to find Gravitational Potential Energy which is mgh. In this case (70.0)(9.81)(60.0) = 41202 J
I don't think...
Homework Statement
A skier of mass 70.0kg is pulled up a slope by a motor-drawn cable. How much work is required to pull the skier 60.0m up a 35 degree slope (assume to be frictionless) at a constant speed of 2.0m/s?
Mass = 70.0kg
Distance = 60.0m
Angle = 35
Velocity/Speed = 2.0m/s
Homework...