A 20-kg block is released from rest at a rough 10.0 m long 30 degree incline. Th

Question

A 20-kg block is released from rest at a rough 10.0 m long 30 degree incline. The coeffient of kinetic friction between incline and box is 0.2. At the bottom of the incline is a frictionless flat track. The block travels down the track, hits a spring of force constant K=100 N/m. Determine (a) the speed of the block at the bottom of the incline, (b) the maximum compressing distance of the spring from its equilibrium position. Please explain clearly and show all work. so i can choose best answer. thanks

Explanation / Answer

at top PE = m*g*h = m*g*L*sin30 = 20*9.8*10*sin30 = 980 J

Wf = u*m*g*cos30*L = 0.2*20*9.8*10*cos30 = 339.48 J

from energy conservation Pe - Wf = KEb

Et - Wf = Eb = 0.5*m*v^2

980-339.48 = 0.5*20*v^2

v = 8 m/s

b) 0.5*m*v^2 = 0.5*k*x^2

20*8*8 = 100*x^2

x = 3.58 m/s

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