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

[pinkyjoshi65]

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..

 

[cyprusad]

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:

$$\frac{1}{2} mv^{2} = \frac{1}{2} kx^{2} - Fs$$

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.

 

[ogb p]

Thank you for providing the context for this scenario. I would approach this problem by first identifying the key variables and equations that are relevant to finding the speed of the BB as it leaves the gun.

The key variables we have are:

- Mass of the projectile (m = 8.24 g = 0.00824 kg)
- Spring constant (k = 2.88 × 103 N/m)
- Length of the barrel (l = 15.1 cm = 0.151 m)
- Frictional force (F = 0.125 N)
- Compression of the spring (x = 16 mm = 0.016 m)

The key equations we can use are:

- Hooke's Law: F = kx, where F is the force applied, k is the spring constant, and x is the displacement of the spring.
- Newton's Second Law: F = ma, where F is the net force, m is the mass of the object, and a is the acceleration.
- Kinematic equation: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.

To find the speed of the BB as it leaves the gun, we need to calculate the acceleration of the projectile as it moves through the barrel. We can do this by using Newton's Second Law and solving for a:

a = F/m = 0.125 N / 0.00824 kg = 15.18 m/s^2

Next, we can use the kinematic equation to find the final velocity of the BB as it leaves the gun. We know that the initial velocity (u) is 0 since the BB starts at rest, and the displacement (s) is the length of the barrel (l = 0.151 m). So, plugging in the values, we get:

v^2 = 0^2 + 2(15.18 m/s^2)(0.151 m) = 4.60 m^2/s^2

Finally, we can take the square root of both sides to find the final velocity:

v = √(4.60 m^2/s^2) = 2.14 m/s

Therefore, the speed of the BB as it leaves the gun is 2.14 m/s.

However, it is important