A 0.24 kg ladle sliding on a horizontal frictionless surface is attached to one

Question

A 0.24 kg ladle sliding on a horizontal frictionless surface is attached to one end of a horizontal spring (k = 300 N/m) whose other end is fixed. The ladle has a kinetic energy of 96 J as it passes through its equilibrium position (the point at which the spring force is zero). At what rate is the spring doing work on the ladle as the ladle passes through its equilibrium position? At what rate is the spring doing work on the ladle when the spring is compressed 0.41 m and the ladle is moving away from the equilibrium position?

Explanation / Answer

Mass = 0.24kg

K=300N/m

Energy E = 96J

Displacement (x) = 0.41

If x<o then conservation of total mechanical energy is

E= 96J = ½ mv2+ ½ kx2 = ½ [ mv2 +kx2]

Spring force does work at the following rate

P =F (dx/dt) = -kxv = -kxsign (v) sqrt ((2E-kx2)/m)

P is power i.e., rate of work done

P= -kx sign (v) sqrt ((2E-kx2)/m)

P= -(300 x (-0.41) x (-1)) sqrt ( ( 2 x 96 -300x (-0.41)2)/0.24)

= -123 sqrt(589.88)

=-123 x24.29

                = -2987.68J/s or (~2.99kJ/s)

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