Answer to Question #228479 in Physics for Tshego

Question

A linear spring has a free length of 41.9cm. When it is under a hanging load of 69506N, it's total length is 866.7cm. 1).Now calculate the spring constant.

2). Calculate the hanging load that makes it 1676.1cm

3). Calculate the energy stored in the spring when the extension is 1676.1cm.

Answer 1

1. From the Hook's law, the spring constant is:

$$k = \dfrac{F}{\Delta l}$$

where $F = 69506N$ is the load and $\Delta l = 8.667m - 0.419m= 8.248m$.

Thus, obtain:

$$k = \dfrac{69506N}{8.248m} \approx 8427N/m$$

2. The load is from the former formula:

$$F = k\Delta l = 8427N/m\cdot (16.761m - 0.419m) \approx 137700N$$

3. The potential energy is given as follows:

$$W = \dfrac{k(\Delta l)^2}{2} = \dfrac{8427\cdot (16.761m - 0.419m)^2}{2} \approx 1125000J$$