a) where mg = mv^2/r
b) 1/2 mv^2
c) KE - Ffd = 1/2 mv^2
Where Ff = mu mg
d) mgh = 1/2 mv^2
You can see m cancels in all occasions.
A block of mass 0.61kg 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 0.40. This section (from points B to C) is 4.73m in length. The block then enters a frictionless loop of radius r= 2.66m. Point D is the highest point in the loop. The loop has a total height of 2r.
a.) What is the minimum speed of the block at point D that still allows the block to complete the loop without leaving the track?
b.) 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?
c.)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?
d.) 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?
3 answers
Yes i can see this
mg=mv^2/r
(0.61)9.8=(v^2(0.61))/2.66
v=2.61
(0.61)9.8=(v^2(0.61))/2.66
v=2.61