Question

A spring with \(k=10.0 \mathrm{~N} / \mathrm{cm}\) is initially stretched \(1.00 \mathrm{~cm}\) from its equilibrium length. a) How much more energy is needed to further stretch the spring to \(5.00 \mathrm{~cm}\) beyond its equilibrium length? b) From this new position, how much energy is needed to compress the spring to \(5.00 \mathrm{~cm}\) shorter than its equilibrium position?

Short answer

Answer: The additional energy needed to stretch the spring to 5.00 cm beyond its equilibrium position is 120.0 Joules. The energy needed to compress the spring to 5.00 cm shorter than its equilibrium position from the new position is 375.0 Joules.

Step-by-step solution

Calculate initial potential energy

First, calculate the initial potential energy (PEi) stored in the spring when it is stretched 1.00 cm from its equilibrium position using the formula \(PE = \frac{1}{2}kx^2\). \(PEi = \frac{1}{2}(10.0 \mathrm{~N} / \mathrm{cm})(1.00 \mathrm{~cm})^2 = 5.00 \mathrm{J}\) The initial potential energy is 5.00 Joules.

Calculate final potential energy for part a

Next, calculate the final potential energy (PEa) stored in the spring when it is stretched to 5.00 cm beyond the equilibrium position using the formula \(PE = \frac{1}{2}kx^2\). \(PEa = \frac{1}{2}(10.0 \mathrm{~N} / \mathrm{cm})(5.00 \mathrm{~cm})^2 = 125.0 \mathrm{J}\) The final potential energy for part a is 125.0 Joules.

Find the additional energy needed for part a

To find the additional energy needed to stretch the spring, subtract the initial potential energy (PEi) from the final potential energy (PEa). Additional energy for part a = \(PEa - PEi = 125.0 \mathrm{J} - 5.00 \mathrm{J} = 120.0 \mathrm{J}\) The additional energy needed to stretch the spring to 5.00 cm beyond its equilibrium position is 120.0 Joules.

Calculate final potential energy for part b

Now, calculate the final potential energy (PEb) stored in the spring when it is compressed 5.00 cm shorter than its equilibrium position from the new position using the formula \(PE = \frac{1}{2}kx^2\). \(PEb = \frac{1}{2}(10.0 \mathrm{~N} / \mathrm{cm})(-10.00 \mathrm{~cm})^2 = 500.0 \mathrm{J}\) The final potential energy for part b is 500.0 Joules.

Find the energy needed for part b

To find the energy needed to compress the spring, subtract the final potential energy (PEa) from the final potential energy (PEb). Energy needed for part b = \(PEb - PEa = 500.0 \mathrm{J} - 125.0 \mathrm{J} = 375.0 \mathrm{J}\) The energy needed to compress the spring to 5.00 cm shorter than its equilibrium position from the new position is 375.0 Joules.