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

I'm not getting exactly your answer but I think this is the right way to do this problem.

I'm assuming that 10 meters from the bottom means 10 meters along the ramp, which at 30 degrees, puts the skater 5 meters above the datum.

This is a combination of using PE and KE transfer on the one part and friction and force for the second part

Given the geometry of the skater down the slide s/he is 5 meters above the datum. (10sin30=5 m)

the speed at the bottom of the first ramp can be found by equating

1/2mv^2=mgh ==> v=sqrt(2gh)=sqrt(2*9.8*5)=9.9 m/s at the bottom of the ramp.

Now for the second part. We'll need to create a free body diagram use the ramp as the frame of reference

In the y direction N=mg cos(25)

in the x direction. Because the body will decelerate the free body equation is

ma=mgsin(25)+umgcos(25). the m's cancel so the deceleration is -5.03 m/s^2. The body comes to rest in t=9.9/5.03=2 s

in that time the body will travel s=1/2(-5.03)(2)^2+2*9.9=9.74 m up the ramp.