Question

PHY 3460 Hooke's Law and Elastic Potential Energy Questions When applying a 37.5 N force on a spring it compresses 15.0 cm. Calculate the spring 2) A spring (k 1.22 N/m) is hanging vertically. An unknown mass is hung from the spring 3) A 15.0 kg mass is hung from a spring causing it to stretch 0.25 m, find the spring constant. constant of the spring. causing it to stretch 57.3 mm. How large is the unknown mass? Then, another 5.0 kg is added to the original mass. How much farther does the spring stretch? 4) How much energy is stored in the spring in problem #1 5) How much energy is stored in the spring in problem #2 6) How much energy is stored in the spring in problem #3 after the 15 kg mass is attached? How much energy is stored in the spring in problem #3 after the 5 kg mass is added? What is the change in energy once the 5kg mass is added? F: forte applicd

Answer 1

1) When the force F is applied on a spring of spring constant k, it is compressed by a length, x. The applied force is balanced by the restoring force of the spring -

$\\|F_{applied}| = |F_{restoring}| \\F_{applied} = kx \\\text{As } x = 15 \text{ cm} = 0.15\text{ m} \\37.5\text{ N} = k(0.15 \text{ m}) \\k = 250 \text{ N/m}$

2) The spring is stretched by the weight of the unkown mass. The gravitational force acting on it is balanced by the restoring force of the spring -

$\\|F_{gravitation}| = |F_{restoring}| \\mg = kx \\m(9.8 \text{ ms}^{-2}) = (1.22 \text{ N/m})(0.0573 \text{ m}) \\m = 7.13 \times10^{-3} \text{ kg} \\m = 7.13 \text{ g}$

3) When the 15 kg mass is hung from a spring, it stretches by 0.25 m, i.e.,

$\\mg = kx \\(15 \text{ kg})(9.8 \text{ ms}^{-2}) = k(0.25 \text{ m}) \\k = 588 \text{ N/m}$

Now when the 5 kg mass is added to the given mass, The total mass becomes 20 kg, therefore,

$\\mg = kx \\(20 \text{ kg})(9.8 \text{ ms}^{-2}) = (588 \text{ N/m})x \\x \approx 0.333 \text{ m}$

Therefore the spring stretches further by (0.333 - 0.25) = 0.083 m

4) The energy stored in a spring is given by -

$\\E = \frac{1}{2}kx^2\\$

Therefore, the energy stored in problem #1 -

$\\E = \frac{1}{2}kx^2\\ \\E = \frac{1}{2}(250)(0.15)^2\\\\E = 2.8125 \text{ J}$

5) The energy stored in problem #2 -

$\\E = \frac{1}{2}kx^2\\ \\E = \frac{1}{2}(1.22)(0.0573)^2\\\\E = 0.002 \text{ J}$

6) Energy after the 15 kg mass is attached -

$\\E = \frac{1}{2}kx^2\\ \\E = \frac{1}{2}(588)(0.25)^2\\\\E = 18.375 \text{ J}$

Energy after the 5 kg mass is added -

$\\E = \frac{1}{2}kx^2\\ \\E = \frac{1}{2}(588)(0.333)^2\\\\E = 32.6 \text{ J}$

The change in energy is given by -

$\\\Delta E = 32.6-18.375 \text{ J} \\\Delta E = 14.225 \text{ J} \\$