What is the downward acceleration of the mass at the point A of the loop? What i

Question

What is the downward acceleration of the mass at the point A of the loop?

What is the minimum height h for which the block will reach point A on the loop without leaving the track?

A small block of mass m = 1.3 kg slides, without friction, along the loop-the-loop track shown. The block starts from the Point P a distance h = 60.0 m above the bottom of the loop of radius R = 20.0 m. What is the kinetic energy of the mass at the point A on the loop?What is the downward acceleration of the mass at the point A of the loop?

Explanation / Answer

(a)KE = PE1 - PE2 = mg(h1-h2) = 1.3*9.8*(60-40) = 254.8 J (b)Velocity = 1/m * sqrt(2KE) = 19.8 m/s (c)For the block to move without leaving the track, centrifugal acceleration should balance weight at the topmost point. mv^2/r = mg v = sqrt(rg) = 14 m/s So, minimum height for reaching this velocity is mgh = mg*(2*20) + 1/2 * m*14^2 h = 50 m

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