A problem about work Hi! I need some help with this question: Jim rides his skateboard down a ramp that is in the shape of a quarter circle with a radius of 5.00 meters. At the bottom of the ramp, Jim is moving at 9.00 m/s. Jim and his skateboard have a mass of 65.0 kg How much work is done by friction as the skateboard goes down the ramp? What I used was W=F(force) x r (displacement) F= mv^2/r = (65 kg x 9 m/s^2)/(5 m) = 1035 N W= (1035 N)(19.6 m) (To find 19.6, I used the circumference equation to find the distance down the ramp) W= 20,286 J 20.3 kJ I really do not think this is correct, and I'm trying to figure out what it means by "frictional" force. Thanks for your help!
Jim and his skateboard have gravitational potential energy at the top of the ramp, what is it? How much kinetic energy would he have if all of his potential energy were transformed into kinetic? How much kinetic energy does Jim have? Remember pe = mgh ke = (mv^2)/2
yes, i think so- but that wouldn't be the same as it being curved, because the displacement on a curved ramp would be different than that on a straight ramp, no?
Not necessarily. If the ramp were a 45 degree incline, the displacement would be the same (5m down, 5m over); however, the way I figured this out does not need a displacement. At the top of the ramp, Jim has no kinetic energy, but he has mgh potential energy. What does he have at the bottom?
mms05 is right, the displacement on a curved ramp would be different than that on a straight ramp, but, in this particular problem you can use a different approach , just like Teegvin said, in order to calculate work. W = - (Uf - Ui) where Uf is the final potential energy and Ui is initial potential energy.
Find the total energy at the top and at the bottom. The difference in energy is the energy wasted to do work against friction when rolling down the ramp. :)