A small block of mass .15kg is placed at a heigh 2m above the bottom of atrack a
ID: 2288240 • Letter: A
Question
A small block of mass .15kg is placed at a heigh 2m above the bottom of atrack and is released from rest. It slides with negligible friction down the track arond the inside loop of radius .6m and leaves the track at a heigh of .5m above the bottom.
A.) Calculate the speed of the block when it leaves the track at point c
B.) Draw and label forces that act on the block when at the top of the loop
C.) Calculate the minimum speed the block can have at point B without losing contact
D.) Calculate the minimum height from which the block can be released and still go around the loop of the track.
Please show me how to work the whole problem out, thanks!
Explanation / Answer
If you use conservation of energy it is completely independent of the path... even if there's a loop.
mgh =.5mv^2
a) 5.42m/s
b) Weight and normal force both acting downwards
c)minimum speed is when centripetal force = gravity or mv^2/r = mg
v^2/r = g
v^2/.6 =9.8
v= 2.42m/s
d) mgh = .5mv^2 where you we set the top of the circle as the reference level. set v equal to 2.42 and solve h... then add that h to 1.2 to get the height above the bottom of the track
1.5m