Question

- One such slide, Der Stuka, named for the terrifying German dive bombers of World War II, is 72.0 feet high (21.9 m), found at Six Flags in Dallas, Texas, and at Wet'n Wild in Orlando, Florida.

(a) Determine the speed of a 59.6-kg woman at the bottom of such a slide, assuming no friction is present. (Assume her initial speed vi = 0.) 20.7 m/s

(b) If the woman is clocked at 16.0 m/s at the bottom of the slide, find the work done on the woman by friction. -5167 J

-Suppose a slide similar to Der Stuka is 36.7 meters high, but is a straight slope, inclined at 45.0° with respect to the horizontal.

(a) Find the speed of a 59.6-kg woman at the bottom of the slide, assuming no friction. ______m/s

(b) If the woman has a speed of 24.5 m/s at the bottom, find the change in mechanical energy due to friction_______. J

(c) Find the magnitude of the force of friction, assumed constant.__________ N

Answer 1

a) speed of the woman at the bottom,

v = sqrt(2*g*h)

= sqrt(2*9.8*36.7)

= 26.8 m/s

b) change in mechanical energy = final kinetic energy - initial potential energy

= (1/2)*m*v^2 - m*g*h

= (1/2)*59.6*24.5^2 - 59.6*9.8*36.7

= -3548 J

c) let fk is magnitude of frictional force acting on the woman.

distance travelled along the incilned surface, d = h/sin(45)

= 36.7/sin(45)

= 51.9 m

now use, Workdone by friction = fk*d*cos(180)

-3548 = -fk*51.9

==> fk = 3548/51.9

= 68.4 N