A block of mass 0.32kg starts from rest at point A and slides down a frictionles

Question

A block of mass 0.32kg 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 0.27. This section (from points B to C) is 5.07m in length. The block then enters a frictionless loop of radius r= 2.07m. Point D is the highest point in the loop. The loop has a total height of 2r. Note that the drawing above 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

to negotiate the loop at the top of the loop
total Fy = mac = mv2 / r
mg - N = mv2 / r
the point at which it would about to fall off the track set N = 0
mg = mv2 / r
g = v2 / r
v = (gr)
v = (9.81 * 2.07)
v = 4.506 m/s

from point c to peak
KE = PE at peak + KE speed needed
KE = mg2r +.5mv2
KE = .32 * 9.81 * 2 * 2.07 +.5 * .32 * 4.5062
KE = 13.00 + 3.249
KE = 16.25 J

from B to C
KE = previous KE + friction's energy
KE = 16.25 + (mg)d
KE = 16.25 + .27 * .32 * 9.81 * 5.07
KE = 16.25 + 4.297
KE = 20.55 J

from A to B
mgh = KE
.32 * 9.81h = 20.55
3.139h = 20.55
h = 6.546 m 

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