A small block slides down a frictionless track whose shape is described by y=x^2
ID: 1517607 • Letter: A
Question
A small block slides down a frictionless track whose shape is described by y=x^2/d for x<0 and y=x^2/d for x>0. The value of d is 0.4 meters, and xand y are measured in meters as usual.
a) Suppose the block starts from rest on the track, at x= 1.4 meters. What will the block's speed be when it reaches x=0? (Answer in m/s)
b) Suppose the block starts on the track at x=0, and is given aninitial velocityof 5.5 m/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 in m)
c) Now suppose the blocks starts on the track at x=3.3 m. 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 beings sliding down. What is its initial velocity in this case? (Answer in 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 in m/s)
e) You start the block on the track at rest, somehwere 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 youre starting to the left of the origin) (Answer in m)
Explanation / Answer
a) the block’s speed = sqrt(2gh)
= sqrt(2 * 9.8 * 1.4 * 1.4 /0.4)
= 9.8 m/sec
b) Here, value of x will the block turn around = v2/(2 * g)
= 5.52/(2 * 9.8)
= 1.543 m
c) initial velocity in this case = sqrt(2 * 9.8 * 3.3 * 3.3 /0.4)
= 23.1 m/sec
d) minimum initial velocity = sqrt(2 * 9.8 * 3.3)
= 8.04 m/sec
e) maximum value of x = - 5.62/(2 * 9.8)
= - 1.6 m