Question

Suppose the roller coaster in Fig. 6-41 (h1 = 32 m, h2 = 12 m, h3 = 25) passes point 1 with a speed of 1.40 m/s. If the average force of friction is equal to one sixth of its weight, with what speed will it reach point 2? The distance traveled is 45.0 m.

Answer 1

Total energy E of the car at point A is the sum of the gravitational potential energy PE and the kinetic energy KE.

PE = m*g*h1
KE = (1/2)m*v1^2
E = m*g*h1 + (1/2)m*v1^2
E = m*[9.81*32 + (1/2)1.4^2] = m*314.901

When the car reaches point B it will have 0 potential energy so the kinetic energy at this point will be the difference between E and the energy lost to friction. The energy lost to friction is just the work performed on the car by friction and this is just the 1/6 weight (m*g) of the car times the distance traveled.

Work = W = force * distance = (m*g/6)*45 = m*(9.81/6)*45 = 73.575*m
Energy lost to fiction = W = 73.575*m

KE = (1/2)m*V^2 = E - W
(1/2)m*V^2 = m*314.901 - 73.575*m
V^2 = 2*(314.901 - 73.575) = 482.562
V = 21.969 m/s

The car will have a speed of about 22 m/s when it reaches point B.

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