A block of mass 0.76kg starts from rest at point A and slides down a frictionles

Question

A block of mass 0.76kg 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 0.29. This section (from points B to C) is 5.38m in length. The block then enters a frictionless loop of radius r= 2.24m. 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.

A) What is the minimum speed of the block at point D that still allows the block to complete the loop without leaving the track?

B) 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?

C) 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?

D) 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?

A block of mass 0.76kg 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 0.29. This section (from points B to C) is 5.38m in length. The block then enters a frictionless loop of radius r= 2.24m. 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. What is the minimum speed of the block at point D that still allows the block to complete the loop without leaving the track? 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? 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? What is the minimum height from which the block should start in order to have enough speed at point D that the block w

Explanation / Answer

A) N = mv^2/r - mg = 0

v = sqrt(rg) = 4.69m/s

b) K.E. at C = 0.76 x 4.69^2 /2 + 0.76 x 2 x2.24g = 41.71 J

c) work done by friction + K.E. at C =K.E. at B

0.29 x 0.76g x 5.38 + 41.71 = K.E. at B

= 53.32 J

d) mgh = 0.76 x 9.8 x h = 53.32

h = 7.16 m

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