A small block slides down a frictionless track whose shape is described by y=x^2

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 2.4 meters, and x and y are measured in meters as usual.

a) Suppose the block starts from rest on the track, at x= -1.7 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 an initial velocityof 1.3 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=1.4 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 begins sliding down. What was 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) by energy conservation,

0.5 mv^2 = mgy

v = sqrt(2*gy)

= sqrt (2*9.8*1.7^2 /2.4)

= 4.86 m/s^2

B) 0.5 mv^2 = mgy

y = 0.5 v^2 /g

x^2 /2.4 =0.5*1.3^2 /9.8

x = - sqrt(0.5*1.3*1.3*2.4/9.8)

= - 0.455 m

C) 0.5 mv^2 = mg|y|

v =sqrt(2gy)

= sqrt(2*9.8*1.4*1.4/2. 4)

= 4 m/s

D) radius of curvature = 1/y" = 1/2*2.4 =1.2 m

mv^2/r = mg

v = sqrt(rg)

= sqrt(1.2*9.8)

= 3.43 m/s answer

E) y = 0.5v^2 /g

x = - sqrt(0.5*3.43*3.43/9.8*2.4)

= - 1.2 m answer

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