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A spring has a natural length of 20 Centimeter. A 40 N force is needed to stretch and hold the spring to a length of 30 Centimeter. How much work is completed in stretching the spring from 35 Centimeter to 38 Centimeter?
Solution : This illustration will need Hooke's Law to find out the force. Law of Hooke tells us that the force needed to stretch a spring a distance of x meters by its natural length is,
F(x) = k x
Here k < 0 is termed as the spring constant.
The first thing that we require to do is find out the spring constant for this spring. We can do which using the initial information. A force of 40 N is needed to stretch the spring 30Centimeter -20Centimeter = 10 Centimeter = 0.10meter from its natural length. By using Law of Hooke we have,
40 = 0.10k ⇒ k = 400
Therefore as per Hooke's Law the force needed to hold this spring x meters from its natural length,
F(x) = 400x
We need to know the work required to stretch the spring from 35Centimeter to 38Centimeter. First we require converting these in distances from the natural length in meters. Doing this provides us x's of 0.15meter and 0.18meter.
The work is now,
us consider the following mass-spring-damper system: md2xdt2+cdxdt+kx=0 with m=5 kg as the mass of the body, k=1.6N/m as the spring constant and two different values of c.
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