Consider the following situation: A block of mass 5 kg is at rest at the top of
ID: 1836934 • Letter: C
Question
Consider the following situation: A block of mass 5 kg is at rest at the top of a 5.6 m ramp. A loop-de-loop, radius of 1 m. is at the bottom of the ramp. After the loop is a 2m long rough patch of track, that has mu_k = 0.3 and mu_s = 0.6, that leads into a 1m ramp that launches the block off the table at an angle of 37 degree from above the horizontal The distance from the table top to the bottom of the pit is 3 m. Assume the whole track is frictionless except the one section of friction. What is the velocity and momentum of the block at point B? Does the block have enough speed at point C to complete the loop-de-loop? What the the energy of the block at point D? What is the velocity of the block when the block leaves the ramp at point E? Right before the block hits the floor, what is the block's velocity? Take the starting energy and the energy the block has right before it hits the floor and verify that the loss of energy equals that due to the rough section of the track.Explanation / Answer
The block has descended a height of 5.6 m
Change in PE = mgh = 5*5.6*9.8 = 274.4 J
As the thre track is friction less no loss of energy and it is equal to the KE of the block
vel v = sqrt(2*274.4/5) = 10.48 m/s
a) momentum = mv = 5*10.48 = 52.4 kg-m/s
b) The PE at the top of the loop (C) = 5*1*9.8 =49 J
The KE of the block at the bottom of the loop is more than the PE required to get to the top of the loop and can complet the loop
c) The only loss of energy in the track is the work done againt friction for 2m length.
frictional force = 5*9.8*0.3 , dynamic friction will come into play as the block is motion
work done agianst friction = 5*9.8*0.3*2= 29.4 J
energy of the block = 274.4 -29.4 = 245 J
d) height of the ramp = 1Sin(37) = 0.6 m
change in PE = 5*9.8*0.6 = 29.4 J
KE at the top of the ramp = 245 -29.4 = 215.6 J
vel v = sqrt(2*215.6/5) = 9.29 m/s
fall of height = 30.6 m
gain in PE = 5*9.8*30.6 = 1499.4 J
Total energy at the bottom of the pit = 1499.4 + 215.6 =1715 J
vel v = sqrt(2*1715/5) = 686 m/s