a 10 lb block is attached to an unstretched spring of constant k = 12 lb/in. The

Question

a 10 lb block is attached to an unstretched spring of constant k = 12 lb/in. The coefficient of static and kinetic friction between the block and the plane are 0.6 and 0.4 respectively. if a force is slowly applied to the block until the tension in the spring reaches 20 lb and then suddenly removed, determine

a.) the velocity of the block as it return to its initial position
b.) the maximum velocity achieved by the block.

 

it won't let me upload the picture, so i'll explain it,

 

the block is attached to the spring and the spring is attached to a wall. The direction of the force is moving horizontally straight pulling the block away from the wall.

Explanation / Answer

Mass of block, m = 10 lb = 4.54 kg Tension in the spring = 20 lb = 89 N The spring constant, K = 12 lb/in = 2100 N/m So the extension in the spring = 89/2100 = 0.042 m The finla compression in the spring = x The kinetic friction force, f = mg = 0.4 * 4.54 * 9.8 = 17.8 N The total energy, (1/2)kx^2 + k(0.042 + x) = (1/2)k(0.042)^2 By solving the above equation we get the final compression in the spring ' x ' So the final expression in the spring, x = 0.025 m a) We have a formula, (1/2)kx^2 = (1/2)mv^2 From the abve equation we get, v = (kx^2/m)^(1/2) = 0.54 m/s So the velocity of block as it returns to its original position = 0.54 m/s b) The total (elongation + compression), x = 0.025 + 0.042 = 0.067 m Then from the known formula,  v = (kx^2/m)^(1/2) = 1.44 m/s So the maximum velocity of the block, v = 1.44 m/s

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