Asked by Olivia
A block starts at rest and slides down a fric- tionless track. It leaves the track horizontally, flies through the air, and subsequently strikes the ground.What is the speed of the ball when it leaves the track? The acceleration of gravity is 9.81 m/s2 .
Answers
Answered by
Damon
You must figure out how far the ball falls vertically (not the distance down the track but that times the sin of the inclination angle of the track from horizontal)
then v^2 = 2 g h
where h is that falling distance in meters
g is gravity acceleration, about 9.8 m/s^2
v is speed in meters.second.
That comes from conservation of energy
(1/2) m v^2 + m g h = constant
h goes down. v goes up
then v^2 = 2 g h
where h is that falling distance in meters
g is gravity acceleration, about 9.8 m/s^2
v is speed in meters.second.
That comes from conservation of energy
(1/2) m v^2 + m g h = constant
h goes down. v goes up
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