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The strings in a compound bow behave approximately like a
spring. A bowstring is pulled back 0.60 m with a maximum force of
267 N (AKA a 60-pound draw weight) to shoot an 18 g arrow.
Assume energy is conserved.
a) At what speed would the arrow be launched? (Note: W = FΔd uses the average
force while F = -kx uses the instantaneous force.)

Respuesta :

ajeigbeibraheem

The speed at which the arrow would be launched is 133.42 m/s

The work-energy theorem asserts that the net work done applied by the forces on a particular object is equivalent to the change in its kinetic energy.

The equation for the work-energy theorem can be computed as:

[tex]\mathbf{W =\Delta K.E}[/tex]

[tex]\mathbf{F\Delta x =\dfrac{1}{2} mv^2}[/tex]

where;

  • Force (F) = 267 N
  • distance Δx = 0.60 m
  • mass (m) = 18 g
  • speed (v) = ???

From the above equation, let make speed(v) the subject of the formula:

∴

[tex]\mathbf{v = \sqrt{\dfrac{2(F \Delta x)}{m}} }[/tex]

[tex]\mathbf{v = \sqrt{\dfrac{2(267 \times 0.60)}{0.018}} }[/tex]

v = 133.42 m/s

Learn more about the work-energy theorem here:

https://brainly.com/question/17081653

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