What must the magnitude of the friction force have been?

Question

The "Niagara Falls" water slide to waterpark and has a height of 76.0 feet (23.2 meters). Because the slide is not completely vertical the total length of the slide is 90.0 feet (27.4 meters). A 70.0 kg person slides down this slide starting from rest.

a) If the slide is frictionless, what should the person's velocity be at the bottom of the slide in meters/second?

b) What if there was friction between the slide and the person. If the velocity of the person at the bottom of the slide was really 38.2 mi/hour how much energy was converted to Thermal Energy in this problem?

c) From what you founded above, What must the magnitude of the friction force have been? (friction force remained the same and constant)

Answer 1

First, let's get the sine of the angle of the slide. As the height of the slide is 23.2 m, and the length of the slide is 27.4 m, the sine of the angle is 23.2/27.4 = 0.847 = sinθ.

a. The velocity of the person at the bottom of the frictionless slide is gotten by

V² = V_o² = 2g sinθ x S V_o is zero, 2 g sinθ is the component of acceleration parallel to the slide, S is the length of the slide.

V² = 2 x 9.8 x 0.847 x 27.4

V = 21.33 m/s

b. First convert 38.2 miles /hr to m/s. The conversion factor is 0.447m/s/1 mi/hr. Multiply

3.82 x 0.447 = 17.08 m/s

Next, calculate the kinetic energy difference between the frictionless and friction cases. This is the energy dissipated as heat.

KE frictionless = 1/2 x70kg x 21.33² = 15924 Joules

KE friction = 1/2 x 70 x 17.08² = 10210 Joules

energy dissipated by friction = 2714 J

c. The net force propelling the person down the slide is F_net = mg x sinθ - F_friction.

This net force is = m x a_net. where a_net = V_friction²/2S = 17.08² /2 x 27.4 = 5.32 m/s². F_net = 372.4

The friction force is mgsinθ - F_net = 70 x 9.8 0.847 - 372.4 = 209 N.