So you think you know Physics???

Question

A skateboard track has the form of a circular arc with a 4.25 m radius, extending to an angle of 90.0° relative to the vertical on either side of the lowest point, as shown in Figure 8-27. A 63.0 kg skateboarder starts from rest at the top of the circular arc.
http://i214.photobucket.com/albums/cc30/bkoz319/08-24alt.gif

(a)What is the normal force exerted on the skateboarder at the bottom of the circular arc?
___kN




Thanks for your time!!!!

Answer 1

sheffieldfan11-

Two force at the bottom: (1) g, and (2) centripetal acceleration. Know g, so lets get centripetal.

Find speed using KE = PE top

PE= mgh = 63kg*9.8*4.25
=2624 J
KE= 2624 J = 1/2 m v^2
v^2 = 2624 J *2/63 kg
v=9.12 m/sec

Centripetal force
F= m*v^2/r

Normal Force:
=mg +m*v^2/r
=m(g+v^2/r)
=63*(9.8+9.12 ^2/4.25)
=1852 N
=1.852 kN

-Remo