Question

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Answer 1

-54 J.

Step-by-step explanation:

The elastic potential energy stored in the spring can be calculated using the formula: Elastic Potential Energy = 0.5 * k * x^2, where k is the spring constant and x is the distance compressed or stretched from the equilibrium position. In this case, k = 240 N/m and x = 0.40 m. Substituting these values into the formula, we get:

Elastic Potential Energy = 0.5 * 240 N/m * (0.40 m)^2 = 19.2 J.

Therefore, the elastic potential energy of the spring is 19.2 J.

The change in elastic potential energy can be calculated using the formula: Change in Elastic Potential Energy = 0.5 * k * (x_f^2 - x_i^2), where k is the spring constant, x_i is the initial compression or stretch distance, and x_f is the final compression or stretch distance. In this case, k = 240 N/m, x_i = 0.40 m, and x_f = 0.30 m. Substituting these values into the formula, we get:

Change in Elastic Potential Energy = 0.5 * 240 N/m * ((0.30 m)^2 - (0.40 m)^2) = -4.56 J.

Therefore, the change in elastic potential energy is -4.56 J.

The elastic potential energy stored in the spring can be calculated using the formula: Elastic Potential Energy = 0.5 * k * x^2, where k is the spring constant and x is the distance compressed or stretched from the equilibrium position. In this case, k = 150 N/m and x = 0.80 m. Substituting these values into the formula, we get:

Elastic Potential Energy = 0.5 * 150 N/m * (0.80 m)^2 = 48 J.

Therefore, the elastic potential energy stored in the spring is 48 J.

The change in elastic potential energy can be calculated using the formula: Change in Elastic Potential Energy = 0.5 * k * (x_f^2 - x_i^2), where k is the spring constant, x_i is the initial compression or stretch distance, and x_f is the final compression or stretch distance. In this case, k = 150 N/m, x_i = 0.80 m, and x_f = -1.00 m. Substituting these values into the formula, we get:

Change in Elastic Potential Energy = 0.5 * 150 N/m * ((-1.00 m)^2 - (0.80 m)^2) = -54 J.

Therefore, the change in elastic potential energy is -54 J.

Hope this Helps! If not correct tell me.

Answer 2

1. The spring is storing

1/2 (240 N/m) (0.40 m)² = 19.2 J ≈ 19 J

of potential energy.

2. When the spring is compressed by 0.30 m, it is storing

1/2 (240 N/m) (0.30 m)² = 10.8 J

so there was a loss of 19.2 J - 10.8 J = 8.4 J.

3. The spring is storing

1/2 (150 N/m) (0.80 m)² = 48 J

of potential energy.

4. Stretching the spring by 0.20 m more has it storing

1/2 (150 N/m) (1.0 m)² = 75 J

so that the extra weight adds 75 J - 48 J = 27 J of energy.