A 0.41 kg ladle sliding on a horizontal frictionless surface is attached to one

Question

A 0.41 kg ladle sliding on a horizontal frictionless surface is attached to one end of a horizontal spring (k = 360 N/m) whose other end is fixed. The ladle has a kinetic energy of 4.1 J as it passes through its equilibrium position (the point at which the spring force is zero). (a) At what rate is the spring doing work on the ladle as the ladle passes through its equilibrium position? (b) At what rate is the spring doing work on the ladle when the spring is compressed 0.030 m and the ladle is moving away from the equilibrium position?

Explanation / Answer

a) at the equilibrium position , spring force is zero

so, rateat which the spring does waork at equilibrium position = zero

b) when the spring is compressed 0.03 m

TE = PE +KE

4.1 = 0.5kx^2 + 0.5mv^2

4.1 = 0.5*360*0.03^2 + 0.5*4.1*v^2

v = 1.385 m/s

Power = F*V

Power = kxV

P = 360*0.03*1.385 = 14.96 W

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