What is the speed of a BB fired from a spring-compressed BB gun?

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The speed of a BB fired from a spring-compressed BB gun can be calculated using the work-energy theorem. Given a spring constant (k) of 2.88 × 103 N/m, a barrel length of 15.1 cm, and a frictional force of 0.125 N, the force exerted by the spring when compressed 16 mm is 4.74 N. The kinetic energy of the BB as it exits the barrel is derived from the potential energy of the spring minus the work done against friction.

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A BB gun fires a 8.24-g projectile by using a compressed spring (k = 2.88 × 103 N/m) and a 15.1 cm long barrel. As the disc travels through the barrel, it experiences a constant frictional force of 0.125 N. If the spring is compressed 16 mm, what is the speed of the BB as it leaves the gun?


So now we know that the Force is kx= 2.88*103*0.016= 4.74N

so..we use F=ma and find a.

then i got stuck..
 
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Gain in kinetic energy during the course of the motion of the disc, from the instant it was fired till the point when it leaves the barrel; is the sum of the potential energy of the compressed spring and the work done by the barrel(which is negative) If v is required, use:

[tex]\frac{1}{2} mv^{2} = \frac{1}{2} kx^{2} - Fs[/tex]

F is the constant retarding force applied by the barrel, and s is the length of the barrel.
The above equation is just the work energy theorem.
 

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