A 4.3 g mass is released from rest at C which has a height of 1 m above the base

Question

A 4.3 g mass is released from rest at C which has a height of 1 m above the base of a loop-the-loop and a radius of 0.22 m. The acceleration of gravity is 9.8 m/s^2. Find the normal force pressing on the track at A, where A is at the same level as the center of the loop. Answer in units of N. Consider a different situation when the initial height at C has not vet been specified. What the minimum kinetic energy of the block at B, which is located at the top of the loop, so that the block can pass by this point without falling off from the track? Answer in unit of J.

Explanation / Answer

Only 1 question at a time pelase

005)
1st we need to find the speed at point A

use:
decrease in potential energy = increase in kinetic energy
m*g*(h1-h2) = 0.5*m*v^2
g*(h1-h2) = 0.5*v^2
9.8*(1-0.22) = 0.5*v^2
9.8*(0.78) = 0.5*v^2
v = 3.91 m/s

Normal force is providing centrepetal acceleration at point A
m = 4.3 g = 4.3*10^-3 Kg
N =m*v^2/r
= (4.3*10^-3)*3.91^2/0.22
=0.299 N

Answer: 0.299 N

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