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.

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

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)

increase masses, decrease separation