An ideal spring is hung vertically from the ceiling. When a 2.0-kg mass hangs at rest from it the spring is extended 6.0 cm from its relaxed length. A downward external force is now applied to the mass to extend the spring an additional 10 cm. While the spring is being extended by the force, the work done by the spring is:
Question
An ideal spring is hung vertically from the ceiling. When a 2.0-kg mass hangs at rest from it the spring is extended 6.0 cm from its relaxed length. A downward external force is now applied to the mass to extend the spring an additional 10 cm. While the spring is being extended by the force, the work done by the spring is:
Solution
To solve this problem, we need to use the formula for the work done by a spring, which is given by:
W = 1/2 * k * (x_f^2 - x_i^2)
where:
- W is the work done by the spring,
- k is the spring constant,
- x_f is the final extension of the spring, and
- x_i is the initial extension of the spring.
First, we need to find the spring constant, k. We know that the weight of the mass extends the spring by 6.0 cm, or 0.06 m. The weight of the mass is given by its mass times the acceleration due to gravity, which is 2.0 kg * 9.8 m/s^2 = 19.6 N. The spring constant is then given by Hooke's law, F = kx, so k = F/x = 19.6 N / 0.06 m = 326.67 N/m.
Next, we need to find the initial and final extensions of the spring. The initial extension, x_i, is the 6.0 cm that the spring is extended by the weight of the mass, or 0.06 m. The final extension, x_f, is this initial extension plus the additional 10 cm, or 0.06 m + 0.10 m = 0.16 m.
Finally, we can substitute these values into the formula for the work done by the spring to find:
W = 1/2 * 326.67 N/m * (0.16 m^2 - 0.06 m^2) = 4.9 J
So, the work done by the spring is 4.9 Joules.
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