Assume the height of the roller coaster is Y = 49 m (take the reference point, Y

Question

Assume the height of the roller coaster is Y = 49 m (take the reference point, Y = 0, as the bottom of the coaster hill). If the car starts from rest at point A. What is the speed of the car at the bottom of the hill? What is the speed of the car when it reaches point B? You measure the speed at point B to be 12 m/s. Is this greater or less than the speed you expected? What may be the reason for the difference? Given the information in part b, What was the net work done on the car from point A to B? What was the work done by gravity on the car from point A to B?

Explanation / Answer

apply energy conservation from top t0 bottom

loss in potential energy =gain in ke

m*g*40 = 1/2*m * v^2

v^2 = 2* 9.8* 40

v = 28 m/s

same here for b

m*g*(40-25) = 1/2*m * v^2

v^2 = 2* 9.8* 15

v = 17.1464 m/s

b)

we got

Vb =17.1464

but measured speed = 12 m/s

reason

energy lost in friction and air drag this is not ideal condition

c)

net work done = change in KE = 1/2* 0.1 * 12^2 = 0.5 *0.1*144 = 7.2 J

work done by gravity = m*g*( 40-25) = 0.1 *9.8 * 15 = 14.7 J

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