Question

In: Physics

A.) A small block slides down a frictionless track whose shape is described by y =...

A.) A small block slides down a frictionless track whose shape is described by y = (x^2) /d for x<0 and by y = -(x^2)/d for x>0. The value of d is 3.27 m, and x and y are measured in meters as usual. Suppose the block starts from rest on the track, at x = -1.07 m. What will the block s speed be when it reaches x = 0?

B.) Same type of track as in the previous problem, this time with d = 3.36 m. The block starts at x = 0, and is given a push to the left with an initial speed of 5.21 m/s, so it starts sliding up the track to the left. At what value of x will the block reverse direction and start sliding back down?

C.) OK, same sort of track, but now with d = 4.50 m. Now suppose the blocks starts on the track at x = 4.18 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 speed in this case?

Solutions

Expert Solution

a) When x = -1.07m, y = 0.35 m [d=3.27 m]

When x = 0 m, y = 0 m

Hence, the potential energy is converted to kinetic energy.

b) When x = 0 m, y = 0 m [d=3.36 m]

Here, the block is given an initial speed of 5.21 m/s to the left. When the block will reach at a height where its velocity will be zero, it will retrace its path.

Hence, kinetic energy will be converted to potential energy.

if y=1.38 m/s x= - (y*d)0.5 = - 2.15 m.

c) when x = 4.18 m, y = - 3.88 m. [d=4.50 m]

To reach at x = 0, the block has to climb to y=0. For this the block needs an initial velocity. Hence, the kinetic energy is converted into potential energy.

to the left


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