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An archer pulls her bowstring back 0.400 m by exerting a force that increases uniformly from zero to 300 N. (a) What is the equivalent spring constant of the bow?N/m(b) How much work is done in pulling the bow?

Respuesta :

Manetho

Answer:

a) 750 N/m

b) 60 J

Explanation:

the elongation of the bowstring x = 0.4 m

force F on full elongation= 300 N

now we know that F=kx

⇒k= F/x

[tex]k= \frac{300}{0.4} = 750 N/m[/tex]

work done in pulling

W= [tex]\frac{1}{2}kx^2[/tex]

W= [tex]\frac{1}{2}750\times0.4^2[/tex]

W=60 J

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