Question

a force of 15n streched a spring to a total length of 30cm. an addnitional force of 10n streched the force spring 5cm further. find the natural length of the spring

Answer 1

Answer on Question#39501, Physics, Mechanics

A force of $15\mathrm{N}$ stretched a spring to a total length of $30\mathrm{cm}$ . An additional force of $10\mathrm{N}$ stretched the force spring $5\mathrm{cm}$ further. Find the natural length of the spring.

Solution:

Consider a force $F$ stretches the spring so that it displaces the equilibrium position by $x$ .

Hooke's Law is a law that shows the relationship between the forces applied to a spring and its elasticity. The relationship is best explained by the equation $F = -kx$ . $F$ is force applied to the spring this can be either the strain or stress that acts upon the spring. $x$ is the displacement of the spring with negative value demonstrating that the displacement of the spring when it is stretched. $K$ is the spring constant and details how stiff the spring is.

Since we know that a $10\mathrm{N}$ force stretches the spring $5\mathrm{cm}$ , the spring constant is

$$k = \frac {F}{x _ {2}} = \frac {1 0}{0 . 0 5} = 2 0 0 \mathrm {N / m}.$$

Use this value of $k$ to find the first extension:

$$x _ {1} = \frac {F}{k} = \frac {1 5}{2 0 0} = 0. 0 7 5 \mathrm {m} = 7. 5 \mathrm {c m}.$$

This caused the spring to stretch to $30~\mathrm{cm}$ , so the natural length is

$$l = x - x _ {1} = 3 0 - 7. 5 = 2 2. 5 \mathrm {c m}.$$

Answer. $l = 22.5$ cm.