A 5.97 kg block free to move on a horizontal, frictionless surface is attached t

Question

A 5.97 kg block free to move on a horizontal, frictionless surface is attached to one end of a light horizontal spring. The other end of the spring is fixed. The spring is compressed 0.130 m from equilibrium and is then released. The speed of the block is 1.17 m/s when it passes the equilibrium position of the spring. The same experiment is now repeated with the frictionless surface replaced by a surface for which (coefficent of kinetic) = 0.345. Determine the speed of the block at the equilibrium position of the spring.

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Explanation / Answer

m =5.67 kg , x =0.114 m , v = 1.07 m/s

From conservation of energy

K1 +U1 = K2 +U2

(1/2)mv^2 +0 = 0+(1/2)kx^2

(5.97*1.17^2) = k(0.130^2)

Spring constant of spring k = 483.57 N/m

Work done by friction = change in mechanical energy

Ff.x =K1 +U1 -K2 -U2

-uk*mg*x = (1/2)mv^2 +0 -0 - (1/2)kx^2

-0.345*5.97*9.8*0.130 = (0.5*5.97*v^2) -(0.5*483.57*0.130^2)

v= 0.699 m/s

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