Question

Suppose that 5 J of work is needed to stretch a spring from its natural length of 36 cm to a length of 51 cm. (a) How much work is needed to stretch the spring from 41 cm to 46 cm? (Round your answer to two decimal places.) 5 Incorrect: Your answer is incorrect. J (b) How far beyond its natural length will a force of 25 N keep the spring stretched? (Round your answer one decimal place.) cm

Answer 1

Answer:

a) 0.6 joules of work are needed to stretch the spring from 41 centimeters to 46 centimeters.

b) The spring must be 5.6 centimeters far from its natural length.

Step-by-step explanation:

a) The work done to stretch the ideal spring from its natural length is defined by the following definition:

$W = \frac{1}{2}\cdot k\cdot (x_{f}-x_{o})^{2}$ (1)

Where:

$k$ - Spring constant, measured in newtons.

$x_{o}$, $x_{f}$ - Initial and final lengths of the spring, measured in meters.

$W$ - Work, measured in joules.

The spring constant is: ($W = 5\,J$, $x_{o} = 0.36\,m$, $x_{f} = 0.51\,m$)

$k = \frac{2\cdot W}{(x_{f}-x_{o})^{2}}$

$k = \frac{2\cdot (5\,J)}{(0.51\,m-0.36\,m)^{2}}$

$k = 444.44\,\frac{N}{m}$

If we know that $k = 444.44\,\frac{N}{m}$, $x_{o} = 0.41\,m$ and $x_{f} = 0.46\,m$, then the work needed is:

$W = \frac{1}{2}\cdot \left(444.44\,\frac{N}{m} \right)\cdot (0.46\,m-0.41\,m)^{2}$

$W = 0.555\,J$

0.6 joules of work are needed to stretch the spring from 41 centimeters to 46 centimeters.

b) The elastic force of the ideal spring ($F$), measured in newtons, is defined by the following formula:

$F = k\cdot \Delta x$ (2)

Where $\Delta x$ is the linear difference from natural length, measured in meters.

If we know that $k = 444.44\,\frac{N}{m}$ and $F = 25\,N$, then the linear difference is:

$\Delta x = \frac{F}{k}$

$\Delta x = \frac{25\,N}{444.44\,\frac{N}{m} }$

$\Delta x = 0.056\,m$

The spring must be 5.6 centimeters far from its natural length.