A skateboarder is practicing on the \"half-pipe\" shown in the figure below, usi

Question

A skateboarder is practicing on the "half-pipe" shown in the figure below, using a special frictionless skateboard. (You can also ignore the kinetic energy of the skateboard's wheels.)

(a) If she starts from rest at the top of the half-pipe, what is her speed at the bottom?

(b) If the skateboarder has mass

what is her apparent weight at the bottom of the half-pipe?

(c) What speed does she then have when she reaches the top edge on the other side of the half-pipe?

(d) Now suppose she has a speed of 15 m/s at the bottom of the half-pipe. What is the highest point she can reach? Hint: This point may be above the edge of the half-pipe. (Enter the distance from the bottom of the half-pipe.)

The roller coaster in the figure below starts with a velocity of 17 m/s. One of the riders is a small girl of mass 35 kg. Find her apparent weight when the roller coaster is at locations Band C. At these two locations, the track is circular, with the radii of curvature given in the figure. The heights at points A, B, and C are

and

Assume friction is negligible and ignore the kinetic energy of the wheels. (The figure is not necessarily drawn to scale.)

weight at B= ?N

Weight at C=?N

Explanation / Answer

a)v=sqrt(2gR)=8.85m/s

b)W=mg+mv^2/r=mg+2mg=3mg=1735N

c)0 by conservation of energy since it is initially 0.

d)Conserving energy,

0.5mv^2=mgh

h=sqrt(v^2/2g)=11.48m

Conserving energy to get vB,

0.5mv^2=mgh+0.5mvB^2

vB=9.64m/s

W=mg-mvB^2/r=17.5N

Similarly, velocity at C vC

0.5mv^2+mgh2=0.5mvC^2

vC=24.94m/s

W=mg+mvC^2/r=1432N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects