A child with a mass of 83 kg is on top of a water slide that is 12 m long and inclined 30 degrees with the horizontal. How fast is the child going at the bottom of the slide (a) if the slide is frictionless? (b) if the coefficient of friction is 0.30?

Answer 1

a. h = 12*s1n30 = 6 m.

V^2 = Vo + 2g*h = 0 + 19.6*6 = 117.6
V = 10.84 m/s.

b. m*g=83kg * 9.8N/kg = 813.4 N.=Wt. of
the child.

Fn = mg*cos30 = 813.4*cos30 = 704.4 N. = Normal = Force perpendicular to the
slide.

Fk = u*Fn = 0.30 * 704.4 = 211.3 N. =
Force of kinetic friction.

PEmax = mg*h-Fk*L = 813.4*6-211.3*12 =
2344.5 Joules.

KE + PE = 2344.5
KE + 0 = 2344.5 At bottom of slide.
KE = 0.5m*V^2 = 2344.5
41.5V^2 = 2344.5
V^2 = 56.5
V = 7.52 m/s.