A small block of mass .15kg is placed at a heigh 2m above the bottom of atrack a
ID: 1703767 • 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.
Explanation / Answer
a) conservation of energy. 2mg=0.5mg+mv^2/2. so v=5.4(m/s). b) at the top of the loop. v=4(m/s). weight. P=1.47(N). centripetal force. F=mv^2/R=4(N). (up) normal force. N=1.47+4=5.47(N) (down) c)it is mv^2/R=mg. so v=sqrt(gR)=2.4(m/s). d) conservation of energy. m*2.4^2/2+mg*1.2=m*g*h so that h=1.5(m)