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A 0.075-kg arrow is fired horizontally. The bowstring exerts a force on the arrow over a distance of 0.90 m. the bow at 40 m/s. What average force does the bow apply to arrow?​

Respuesta :

yerajg7676

Answer:

Use work-energy to solve this

The work done by on the bowstring will be converted into kinetic energy of the arrow

so W = F*d = K = 1/2m*v^2

o v = sqrt(2*F*d/m) = sqrt(2*65N*0.90m/0.075kg) = 39.5 m/s

Explanation:

snehashish65

The magnitude of average force applied by the bow to the arrow is 66.67 N.

Given data:

The mass of arrow is, m = 0.075 kg.

The distance covered by the arrow is,  s = 0.90 m.

The speed of bow is, v = 40 m/s.

In this problem, the work - energy theorem can be applied. As per the work -energy theorem, the work done by the applied force is equal to the kinetic energy change.

[tex]W = \Delta KE\\\\F \times s = \dfrac{1}{2}mv^{2}[/tex]

Here, F is the magnitude of average force.

Solving as,

[tex]F \times 0.90 = \dfrac{1}{2} \times 0.075 \times 40^{2}\\\\F = 66.67 \;\rm N[/tex]

Thus, we can conclude that the magnitude of average force applied by the bow to the arrow is 66.67 N.

Learn more about the average force here:

https://brainly.com/question/19386420

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