The 96.6 lb block slides down a smooth quarter-circle track of radius R=5 ft and

Question

The 96.6 lb block slides down a smooth quarter-circle track of radius R=5 ft and hits a spring of constant k= 64.4 lb/ft . If the coefficient of kinetic friction of the horizontal surface is 0.25 , determine the maximum deformation of the spring when it arrests the block temporarily.

The 96.6 lb block slides down a smooth quarter-circle track of radius R=5 ft and hits a spring of constant k= 64.4 lb/ft . If the coefficient of kinetic friction of the horizontal surface is 0.25 , determine the maximum deformation of the spring when it arrests the block temporarily.

Explanation / Answer

loss of potential energy = spring potential energy+friction work ;

mgR = 0.5*k*x^2 + 0.25*mg*x ;

483 = 32.2 x^2 +24.15 x ;

x=3.516 , -4.2666 ..

negative x doesnt make sense so x=3.516 ft

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