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

[bikerboi92]

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?

 

[w3390]

(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.

 

[astrokreng]

[/B]

The rate at which the spring is doing work on the ladle can be calculated using the formula for work done by a spring, which is W = 1/2*k*x^2, where k is the spring constant and x is the distance the spring is compressed or stretched. In this case, the ladle is moving away from the equilibrium position, so the spring is being stretched.

We can use the given information to calculate the distance the spring is being compressed. Since the ladle has a kinetic energy of 10 J when passing through the equilibrium position, we can use the formula for kinetic energy, KE = 1/2*m*v^2, where m is the mass of the ladle and v is its velocity, to find the velocity of the ladle at the equilibrium position.

10 J = 1/2 * 0.23 kg * v^2

v = √(10 J * 2 / 0.23 kg) = 6.47 m/s

Now, using the conservation of energy principle, we can find the maximum compression distance of the spring by equating the kinetic energy of the ladle at the equilibrium position to the potential energy stored in the spring when compressed.

1/2 * 0.23 kg * (6.47 m/s)^2 = 1/2 * 490 N/m * x^2

x = √(0.23 kg * (6.47 m/s)^2 / 490 N/m) = 0.10 m

Therefore, when the spring is compressed 0.10 m and the ladle is moving away from the equilibrium position, the spring is doing work on the ladle at a rate of:

W = 1/2 * 490 N/m * (0.10 m)^2 = 2.45 J/s = 2.45 watts

Therefore, the spring is doing work on the ladle at a rate of 2.45 watts when the spring is compressed 0.10 m and the ladle is moving away from the equilibrium position.