Hi please help, (dynamics) cant figure it out?? have the answer , just cant do it?

Question

a skateboarder slides down a frictionless ramp inclined at an angle 30degrees to the horizontal.he then slides across a frictionless horizontal floor and begins to slide up a second incline at an angle of 25degrees to the horizontal.the skateboarder starts at a distance of 10m from the bottom of the first incline.how far up the second incline will he go if the coefficient of kinetic friction on the second incline is 0.10?

the answer is 9.8 metres

thanks!!!!

Answer 1

you're missing some significant information like how long the initial ramp is, whether he started from rest or not, ect...

Basically you need to figure out how much energy the boarder has when he hits the upper incline and then use work to figure out how far up the ramp he goes.

Answer 2

Don't have the time to do your problem, but here's the physics.

The kinetic energy gained in dropping down the first ramp = the energy lost to potential energy plus lost to friction going up the second ramp. Because the horizontal travel between ramps is frictionless, it has no bearing on the energy levels.

The kinetic to potential and vice versa changes are based on vertical displacement only. The friction energy loss is based on distance traveled on the second ramp (the hypoteneuse of the vertical and horizontal displacements on the second ramp).

PS: Got time now. To put what I said above in math talk:

PE down = PE up + friction work up
mgH = mgh + kmg cos(theta) d; where m = mass of skateboarder, g = 9.81 m/sec^2, k = .10, d = distance slid on second ramp (the answer), theta = 25 deg.

We can simplify this to:

H = h + kcos(theta)d; where H = S sin(phi), h = d sin(theta), S = 10 m, the starting distance on frictionless ramp one, phi = 30 deg incline. Therefore: H = h + k cos(theta) h/sin(theta) = h(1 + k cot(theta)) = 10 sin(30) = h(1 + .10 cot(25)).

Solve for h and put that answer into d = h/sin(theta) and, there you have it, the distance the skateboarder will travel up the second ramp.