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Consider a loop-the-loop system where the radius of the loop is r. a small block (of mass m and negligible size) is released from rest at the point p, which is at a height of 41 3 r.

Respuesta :

shinmin
The answer for this problem would be computed using the law of energy conservation which is:=>KE(A) = PE(P)

=>1/2mv^2 = mgh 

=>v^2 = 2gh = 2 x g x 17R/8 = 17gR/4 ----------(i) 

By N(A) = Mv^2/R + Mg 

=>N = M17gR/4R + Mg 

=>N = MG(17/4 + 1) = 21/4 MG

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