A block of mass 0.42kg starts from rest at point A and slides down a frictionles

Question

A block of mass 0.42kg 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 0.26. This section (from points B to C) is 4.05m in length. The block then enters a frictionless loop of radius r= 1.72m. 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

here is the correct answer with correct numerical values a)in a vertical loop, the particle speed should be sqrt(g*r) at the top point v1 =sqrt(g*r)=4.10561 m/s b)in a vertical loop, the particle speed should be sqrt(5*g*r) at the bottom v2 =sqrt(5*g*r)=9.18041 m/s KE2 = 1/2*m*v2^2 = 17.6988 J..... c)KE3 = 1/2 *m*v3^3 = ((1/2)*m*v2^2)+( mue *m*g*s) = 22.03295 J d) KE3 = m*g*h====> h =KE3/(g*m)= 5.353 m

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