A block of mass 0.48kg starts from rest at point A and slides down a frictionles

Question

A block of mass 0.48kg 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 0.30. This section (from points B to C) is 3.71m in length. The block then enters a frictionless loop of radius r= 3.78m. 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

g=v^2/r

v=sqrt(g*r)
6.086 m/s

at C
KE=m*g*2*r+.5*m*6.086^2
124.45J

At B
.5*m*v^2=124.45+m*g*.3*3.71
solve for v
v=sqrt(2*(124.45/m+g*.3*3.71))
23.009m/s

h
m*g*h=.5*m*23.009^2
solve for h
h=.5*23.009^2/9.81
h=26.9m

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