A ball is released from rest at t=0.3s on an inclined segement of track. The bal

Question

A ball is released from rest at t=0.3s on an inclined segement of track. The ball rolls down the incline onto a 2m section of level track and then continues down a second incline, which is inclined at the same angle as the first section of track (and therefore causes the same acceleration). At the end of the first incline t=2.3s. At the end of the straight part of the track, t=2.7s. At the end of the second incline, t=3.1s.
A) What is the speed of the ball on the level segment of track? Show your work.
B) Determine the numerical value of the acceleration of the ball on the first inclined segment of track. Include a description of the direction of the acceleration.
C) Determine the final speed of the ball at time t=3.1s. (Hint: Find out how much the speed increases in the given time.

Explanation / Answer

lenght of incline segment = 2m then inclination angel = theta 2 = (4.9 *2*2*sin theta) angel = 5.857 degrees a) conservation of energy then (4.9)(2 sin theta ) = 0.5* (v^2) v = 1.414 m/s b) acceleration = gsin theta downwards along incline = 9.8 sin5.857 = 1.00005072 m/s^2 c)'"zero "'since according to law of conservation of energy

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