A wooden block with mass 0.5kg is forced aginst horizonatal spring whose elastic

Question

A wooden block with mass 0.5kg is forced aginst horizonatal spring whose elastic constant is 120 N/m compressing the spring a distance of 0.20 m. When released the block moves on a horizontal tabletop for 0.90 m before the block coming to rest (a) What is the coefficient of kinetic friction same spring causing the same compression as in part (a) but now the block is released near the edge of the tabletop. If the tabletop is located 1.2 m above the floor find the distance from the foot of the table at which the block will land and the speed with which it lands

Explanation / Answer

a) By energy conservation,

Spring potential energy = energy lost by friction

0.5kx^2 = umgd

0.5*120*0.2*0.2 = u*0.5*9.8*0.9

u = 0.5*120*0.2*0.2 /(0.5*9.8*0.9)

= 0.544

b) time of flight t is given by

h = 0.5at^2

t = sqrt(2h/a) =sqrt(2*1.2/9.8) = 0.4949 s

Spring energy = KE at release

0.5*120*0.2*0.2 = 0.5*0.5*v^2

v = sqrt(120*0.2*0.2/0.5) = 3.098 m/s

horizontal distance = vt = 3.098 m/s *0.4949 s

= 1.533 m

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