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 0.31. This section (from points B to C) is 3.52m in length. The block then enters a frictionless loop of radius r= 2.52m. 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 point D. (Applying Fnet = ma)

N + mg = mv^2 / r

N = mv^2/r - mg

to keep in contact, N >= 0

for minimum case

N = mv^2/r - mg = 0

v = sqrt(rg) = sqrt(2.52 x 9.8) = 4.97 m/s ....Speed at D

2.Now using energy conservation for C and D

PE + KE at C = PE + KE at D

0 + KE = m g (2r) + m (4.97^2) /2

KE = (0.76 x 9.8 x 2 x 2.52 ) + ( 0.76 x 4.97^2 /2 ) = 46.92 J

3.

For B to C.

Using work energy theorem,

work done by friction = change in KE

- uk m g d = KE at C - KE at B

- (0.31 x 0.76 x 9.8 x 3.52 ) = 46.92 - Kb

Kb = 55.03 J

4. Now using energy conservation for A to B .

mgh = Kb

0.76 x 9.8 x h = 55.03

h = 7.4 m

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