A 0.22 kg ladle sliding on a horizontal frictionless surface is attached to one

Question

A 0.22 kg ladle sliding on a horizontal frictionless surface is attached to one end of a horizontal spring (k = 508 N/m) whose other end is fixed. The ladle has a kinetic energy of 10 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? W (b) At what rate is the spring doing work on the ladle when the spring is compressed 0.10 m and the ladle is moving away from the equilibrium position? W

Explanation / Answer

a) spring force at the equilibrium is zero

hence the spring does not do any work on mass

so rate of work done is P = 0 W

b) rate of work done is P = F*v

F is the spring force

F = k*x

F = 508*0.1 = 50.8 N

at 0.1 m,speed is V

using law of conservation of energy

0.5*m*v^2 + 0.5*k*x^2 = 10

(0.5*0.22*v^2)+(0.5*508*0.1^2) = 10

v = 8.23 m/sec

then

Power is P = F*v = 50.8*8.23 = 418.084 W

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