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:
[tex]W = \frac{1}{2}\cdot k\cdot (x_{f}-x_{o})^{2}[/tex] (1)
Where:
[tex]k[/tex] - Spring constant, measured in newtons.
[tex]x_{o}[/tex], [tex]x_{f}[/tex] - Initial and final lengths of the spring, measured in meters.
[tex]W[/tex] - Work, measured in joules.
The spring constant is: ([tex]W = 5\,J[/tex], [tex]x_{o} = 0.36\,m[/tex], [tex]x_{f} = 0.51\,m[/tex])
[tex]k = \frac{2\cdot W}{(x_{f}-x_{o})^{2}}[/tex]
[tex]k = \frac{2\cdot (5\,J)}{(0.51\,m-0.36\,m)^{2}}[/tex]
[tex]k = 444.44\,\frac{N}{m}[/tex]
If we know that [tex]k = 444.44\,\frac{N}{m}[/tex], [tex]x_{o} = 0.41\,m[/tex] and [tex]x_{f} = 0.46\,m[/tex], then the work needed is:
[tex]W = \frac{1}{2}\cdot \left(444.44\,\frac{N}{m} \right)\cdot (0.46\,m-0.41\,m)^{2}[/tex]
[tex]W = 0.555\,J[/tex]
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 ([tex]F[/tex]), measured in newtons, is defined by the following formula:
[tex]F = k\cdot \Delta x[/tex] (2)
Where [tex]\Delta x[/tex] is the linear difference from natural length, measured in meters.
If we know that [tex]k = 444.44\,\frac{N}{m}[/tex] and [tex]F = 25\,N[/tex], then the linear difference is:
[tex]\Delta x = \frac{F}{k}[/tex]
[tex]\Delta x = \frac{25\,N}{444.44\,\frac{N}{m} }[/tex]
[tex]\Delta x = 0.056\,m[/tex]
The spring must be 5.6 centimeters far from its natural length.
