A small block slides down a frictionless track whose shape is described by y=x2/
ID: 1512538 • Letter: A
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
y = 2x^2 for x<0 and y = -2x^2 for x>0.
a.) x=-3.2 => y = 2(3.2)^2 = 20.48 m
Block starts at y = 20.48 m and reaches y=0, while attaining a velocity v.
Law of conservation of energy:
Kinetic energy gained = potential energy lost
=> 0.5 mv^2 = mgh
=> v = sqrt(2*g*h)
= sqrt(2*9.8*20.48)
v = 20.03 m/s
b.) x = 0. v = 9.1 m/s. Let block reach height y.
Law of conservation of energy:
Kinetic energy gained = potential energy lost
=> 0.5m(9.1)^2 = mgy
=> y=0.5*82.8 / 9.8
=> y = 4.225 m
x = sqrt(y/2) = 1.45 m
c.) x = 3.1 => y = -2(3.1)^2 = -19.22 m
Block starts at y=- 19.22 m with a velocity v and reaches y=0 with a velocity 0.
Law of conservation of energy:
Kinetic energy lost by losing velocity from v to 0 = potential energy gained by increased height by 4.5 meters
0.5mv^2 = mg*19.22
v = sqrt(2*g*y)
= sqrt(2*9.8*19.22)
v = 19.41 m/s
d.) Let v be the velocity at which block attains free-fall, when pushed.
For free fall, at any time, the y distance covered by the block must be less than the y distance of the platform.
At time t, x-coordinate of block = vt
At time t, y-coordinate of block = 0.5gt^2
At x=vt, y-coordinate of platform = -2x^2 = -2v^2t^2
For free fall,
|-2v^2t^2| - |0.5gt^2| > 0
or 2v^2-0.5g > 0
v^2 > g/4
v > root(g/4) = 1.565 m/s.
v > 1.565 m/s
min velocity = 1.565 m/s
e.)
The body is in freefall, if at every point of time, it doesn't touch the platform, and falls freely.
The block must be dropped from a height y such that when it loses potential energy worth mg*y, it must gain kinetic energy equivalent to min velocity required for freefall computed above = 1.565 m/s
when left from height y, block when leaves with freefall => velocity v at y=0 is 1.565 m/s
Law of conservation of energy:
Kinetic energy gained = potential energy lost
0.5mv^2 = m*g*y
=> y = v^2/2g
= 1.565^2/(2*9.8)
= 0.13 m.
Now we have y, so we substitute it in the equation of platform to get x.
x = root(y/2)
= root(0.13/2)
= 0.25
min x = 0.25 m