Question

Suppose the roller-coaster car in fig.6–41 passes point 1 with a speed of if the average force of friction is equal to 0.23 of its weight, with what speed will it reach point 2? the distance traveled is 45.0 m.

Answer 1

The values in figure 1 shows h1 = 39 m, h2 = 13 m, h3 = 25 with a speed of 1.5m/s

 

The Initial speed u = 1.5m/s.

Vertical distance covered between point 1 and point 2 is (39-13) =26m

The expanse 45 m is equal to a height of (45/6) =7.5 (As frictional force is mg/6)

With the kinematics equation v^2 =u^2 + 2as

V^2 = 1.5^2 + 2*9.8 (26 + 7.5)

V = 25.7 m/s