Question

Suppose the roller coaster in fig. 6-41 (h1 = 32 m, h2 = 15 m, h3 = 20) passes point 1 with a speed of 1.80 m/s. if the average force of friction is equal to one sixth of its weight, with what speed will it reach point 2.

A)2.40 m/s

B)2.00 m/s

C)1.50 m/s

D) 1.20 m/s

Answer 1

The final speed of the roller coaster at point 2 is 2.40 m/s.

To find the final speed of the roller coaster at point 2, we can use the principle of conservation of mechanical energy. When friction is negligible, the roller coaster's initial potential energy at point 1 is equal to its final kinetic energy at point 2.

The potential energy at point 1 can be calculated using the equation PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. The kinetic energy at point 2 can be calculated using the equation KE = (1/2)mv^2, where v is the final speed.

Setting the potential energy equal to the kinetic energy, we have mgh1 = (1/2)mv^2. We can then cancel out the mass and solve for v:

gh1 = (1/2)v^2
v^2 = 2gh1
v = sqrt(2gh1)

Plugging in the values h1 = 32 m and g = 9.8 m/s^2, we get v = sqrt(2 * 9.8 * 32) = 2.40 m/s.