A small block of mass m = 1.1 kg slides, without friction, along the loop-the-lo
ID: 2220434 • Letter: A
Question
A small block of mass m = 1.1 kg slides, without friction, along the loop-the-loop track shown. the block starts from the point P a distance h = 53.0 m above the bottom of the loop of radius R = 16.0 m.What is the downward acceleration of the mass at the point A, the apex of the loop?
I have solved for KE and got 226.38 J, which is correct.
Using that, I solved for velocity by KE = 1/2 mv^2 yielding a velocity of 20.27 m/s
Then, I used my velocity to determine the centripetal acceleration by ac = v^2/r and got 25.68 m/s^2
I added the centripetal acceleration to the acceleration due to gravity and got a total downward acceleration of 35.48 m/s^2. Unfortunately this answer is incorrect and I have no idea what to do from here. Please help!
Explanation / Answer
(0.5mv^2 = mgh) (v = sqrt{2g 21} = 20.29m/s) (a_{net} = g+v^2/r = 35.53m/s^2)