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A toy gun uses a spring to project a 5.3 - g soft rubber sphere horizontally. The spring constant is 8.0 N/m, the barrel of the gun is 15 cm long, and a constant frictional force of 0.032 N exists between barrel and projectile. With what speed does the projectile leave the barrel if the spring was compressed 5.0 cm for this launch?

Respuesta :

Answer:

v = 1.16 m/s

Explanation:

As per work energy theorem we know that work done by all forces on the sphere is equal to the change in kinetic energy of the sphere

So here we will have

[tex]\frac{1}{2}kx^2 - F_f .d = \frac{1}{2}mv^2[/tex]

here we know that

[tex]F_f = 0.032 N[/tex]

k = 8 N/m

now we have

[tex]\frac{1}{2}(8)(0.05^2) - (0.032)(0.15 + 0.05) = \frac{1}{2}(0.0053)v^2[/tex]

[tex]0.01 - 6.4 \times 10^{-3} = 2.65 \times 10^{-3} v^2[/tex]

[tex]v = 1.16 m/s[/tex]

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