A spring of force constant 300.0 N/m and unstretched length 0.240 m is stretched by two forces, pulling in opposite directions at opposite ends of the spring, that increase to 15.0 N. How long will the spring now be, and how much work was required to stretch it that distance

Respuesta :

Answer:

[tex] F = K \Delta L[/tex]

And if we solve for the change of length we got:

[tex] \Delta L = \frac{F}{K} =\frac{15N}{300 N/m}= 0.05 m[/tex]

And then the final length would be

[tex] L_f = L_i + \Delta L = 0.24 +0.05 - 0.29 m[/tex]

So then we can find the work required to strech the spring with the following formula:

[tex] W = \frac{1}{2} K (\Delta L)^2 =\frac{1}{2} (300 N/m) (0.05 m)^2 = 0.375 J[/tex]

Explanation:

We can't use the definition of work given by:

[tex] W = F \Delta L[/tex]

Becuase the force in the last formula needs to be constant, and for this case the force associated to the spring is not constant during the movement.

For this case we have the following data given:

[tex] K = 300 N/m[/tex] the constant of the spring

[tex] L = 0.24 m[/tex] the length for the unstreched length

[tex] F = 15N [/tex] represent the force acting

Since we have a change in the length we can write the Hooke's law like this:

[tex] F = K \Delta L[/tex]

And if we solve for the change of length we got:

[tex] \Delta L = \frac{F}{K} =\frac{15N}{300 N/m}= 0.05 m[/tex]

And then the final length would be

[tex] L_f = L_i + \Delta L = 0.24 +0.05 - 0.29 m[/tex]

So then we can find the work required to strech the spring with the following formula:

[tex] W = \frac{1}{2} K (\Delta L)^2 =\frac{1}{2} (300 N/m) (0.05 m)^2 = 0.375 J[/tex]

The final length of the spring is 0.29 m and the work required to stretch it at that distance is 0.375 J

GIven that:

  • the spring force constant k = 300 N/m
  • length (x) = 0.240 m
  • Force increase to 15.0 N

The stretch in the length of the spring [tex]\mathbf{ \Delta x =\dfrac{15.0 N}{300 \ N/m}}[/tex]

= 0.05 m

The final length of the spring after pulling the force in opposite direction

= 0.240 m + 0.05 m

= 0.29 m

The work required to stretch the spring at a distance of 0.29 m can be calculated by using the Hooke's Law of elasticity;

F = k× Δx

[tex]\mathbf{W = \dfrac{1}{2} \times k \times \Delta x^2}[/tex]

[tex]\mathbf{W = \dfrac{1}{2} \times 300 N/m\times (0.05)^2}[/tex]

[tex]\mathbf{W =0.375 \ J}[/tex]

Therefore, we can conclude that the final length of the spring is 0.29 m and the work required to stretch it at that distance is 0.375 J

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