A toy gun uses a spring to project a 5.6 g soft rubber sphere horizontally. The spring constant is 10.0 N/m, the barrel of the gun is 14.3 cm long, and a constant frictional force of 0.036 N exists between barrel and projectile. With what speed does the projectile leave the barrel if the spring was compressed 6.7 cm for this launch? (Assume the projectile is in contact with the barrel for the full 14.3 cm.)
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To find the speed with which the projectile leaves the barrel, we can use the principle of conservation of energy.
The potential energy stored in the compressed spring is given by the equation:
E_potential = (1/2) * k * x^2
where k is the spring constant and x is the displacement of the spring (in this case, the compression of 6.7 cm, or 0.067 m).
Given that the spring constant is 10.0 N/m and the displacement is 0.067 m, we can calculate the potential energy:
E_potential = (1/2) * 10.0 N/m * (0.067 m)^2
= 0.022755 N
The work done by the frictional force is equal to the force of friction multiplied by the distance, which is 14.3 cm or 0.143 m:
Work_friction = force_friction * distance
= 0.036 N * 0.143 m
= 0.005148 N
The total mechanical energy of the system is conserved, so the potential energy stored in the spring is converted into the kinetic energy of the projectile:
E_total = E_kinetic + E_potential + Work_friction
where E_total is the total mechanical energy, E_kinetic is the kinetic energy of the projectile, and Work_friction is the work done by friction.
Since the projectile starts from rest and leaves the barrel horizontally, its initial kinetic energy is zero. Therefore, we can simplify the equation to:
E_total = E_potential + Work_friction
E_total = 0.022755 N + 0.005148 N
= 0.027903 N
The total mechanical energy is also equal to the final kinetic energy of the projectile:
E_total = (1/2) * m * v^2
where m is the mass of the projectile (5.6 g or 0.0056 kg) and v is its velocity.
Solving for v:
v = sqrt((2 * E_total) / m)
= sqrt((2 * 0.027903 N) / 0.0056 kg)
≈ sqrt(0.0996542 m^2/s^2)
≈ 0.315 m/s
Therefore, the projectile leaves the barrel with a speed of approximately 0.315 m/s.