How Much Work is Done by a Spring on a Sliding Ladle?

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The discussion focuses on the work done by a spring on a ladle sliding on a frictionless surface. At the equilibrium position, the spring does no work on the ladle, resulting in a power output of 0 watts. When the spring is compressed by 0.10 m, the force exerted by the spring is calculated to be 49 N. The work done by the spring at this compression, as the ladle moves away from the equilibrium, is determined to be 4.9 J, though it may be negative due to the direction of force. Understanding the relationship between force, distance, and work is crucial in this context.
bikerboi92
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Ladle Attached to Spring A 0.23 kg ladle sliding on a horizontal frictionless surface is attached to one end of a horizontal spring (with k = 490 N/m) whose other end is fixed. The mass 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?

i got this correct: 0 watts

(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?
 
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bikerboi92 said:
(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?

First you must find the force in the spring at x=.1m which can be found through the equation F=kx. Therefore, the force will be 49N. Then you know that WORK=FORCE*DISTANCE. Therefore W=(49N)*(.1m)=4.9J. This should be correct, though it might be negative.
 

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