A block of mass 0.79kg starts from rest at point A and slides down a frictionles

Question

A block of mass 0.79kg starts from rest at point A and slides down a frictionless hill of height h. At the bottom of the hill it slides across a horizontal piece of track where the coefficient of kinetic friction is 0.48. This section (from points B to C) is 5.14m in length. The block then enters a frictionless loop of radius r= 2.19m. Point D is the highest point in the loop. The loop has a total height of 2r. Note that the drawing below is not to scale.
1. What is the minimum speed of the block at point D that still allows the block to complete the loop without leaving the track?
2. What is the minimum kinetic energy for the block at point C in order to have enough speed at point D that the block will not leave the track?
3. What is the minimum kinetic energy for the block at point B in order to have enough speed at point D that the block will not leave the track?

4. What is the minimum height from which the block should start in order to have enough speed at point D that the block will not leave the track?

Explanation / Answer

1. at highest point ,

N + mg = mv^2 /r

N   = mv^2 / r - mg   for minimum velocity casse N = 0

mv^2 / r - mg = 0

v^2 = rg

v =sqrt(9.81 x 2.19) = 4.64 m/s

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2. using work-energy theroem ,

work done by gravity = change in K.E.

- mg(2r) = m (4.64)^2 /2    - mv^2 /2

v^2 /2 = 4.64^2 /2 + 9.81*2*2.19

v = 10.37 m/s

K.E. = mv^2 /2   = 0.79 x 10.37^2 /2 = 42.45 J

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3. agian using work -energy theorem ,

work done by friction = change in K.E.

- umgd = m (10.37)^2 /2 - K.E.

K.E. = m ( 10.37^2 /2   + 0.48x9.81 x 5.14) = 77.97x0.79 = 61.60 J

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4.

K.E. = mgh

61.60 = 0.79 x 9.81 x h

h   = 7.95 m

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