a small block slides down a frictionless track ... Your question has been answer

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a small block slides down a frictionless track ... Your question has been answered! Rate it below. Let us know if you got a helpful answer. Question A small block slides down a frictionless track whose shape is described by y=x2/d for x<0 and y=-x2/d for x>0. The value of d is 0.8 meters, and x and y are measured in meters as usual.

(a) Suppose the block starts from rest on the track, at x = -3.2meters. What will the block’s speed be when it reaches x = 0? Answer: ______________.___ m/s

( b) Suppose the block starts on the track at x = 0, and is given an initial velocity of 9.1m/s to the left. The block then begins to slide up the track to the left. At what value of x will the block turn around and begin to slide down again? Answer: ___________.___ m

(c) Now suppose the blocks starts on the track at x = 3.1m. The block is given a push to the left and begins to slide up the track, eventually reaching its maximum height at x = 0, at which point it turns around and begins sliding down. What was its initial velocity in this case? Answer: ____________.___ m/s

(d) Suppose the block starts on the track at x = 0. What minimum initial velocity (moving to the right) must the block have such that it will leave the track at x = 0 and go into freefall? Answer: ____________.____ m/s

(e) You start the block on the track at rest, somewhere to the left of x = 0. You then release the block from rest and let it slide down. What is the maximum value of x from which you can release the block from rest and have it leave the track at x = 0 and go into freefall? (Note: your answer should be a negative number, since you’re starting to the left of the origin.) Answer: ____________.___ m

Explanation / Answer

a) the block’s speed = sqrt(2gh)

                                  = sqrt(2 * 9.8 * 3.2 * 3.2 /0.8)

                                  = 15.839 m/sec

b)    Here,    value of x will the block turn around = v2/(2 * g)

                                                 = 9.12/(2 * 9.8)

                                                 = 4.225 m

c)    initial velocity in this case = sqrt(2 * 9.8 * 3.1 * 3.1 /0.8)

                                                = 15.34 m/sec

d)    minimum initial velocity =   sqrt(2 * 9.8 * 3.1)

                                                =   7.79 m/sec

e)     maximum value of x   =   - 9.22/(2 * 9.8)

                                          =   - 4.318 m

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