A toy gun uses a spring to project a 6.1 g soft rubber sphere horizontally. The spring constant is 12.0 N/m, the barrel of the gun is 12.6 cm long, and a constant frictional force of 0.031 N exists between barrel and projectile. With what speed does the projectile leave the barrel if the spring was compressed 6.3 cm for this launch? (Assume the projectile is in contact with the barrel for the full 12.6 cm.)

Well, sounds like we have a spring-loaded rubber sphere shooter on our hands! Let's calculate the speed at which this rubber sphere will leave the barrel, shall we?

First, let's find the work done by the spring. The formula for the work done by a spring is:

W = (1/2)kx^2

Where W is the work done, k is the spring constant, and x is the displacement of the spring. In this case, the spring constant is 12.0 N/m and the displacement is 6.3 cm (which can be converted to meters as 0.063 m).

So, the work done by the spring is:

W = (1/2)(12.0)(0.063)^2

Now, let's find the work done against friction. The formula for work done against friction is:

W = Fd

Where W is the work done, F is the frictional force, and d is the distance traveled against friction. In this case, the frictional force is 0.031 N and the distance traveled against friction is 12.6 cm (which can be converted to meters as 0.126 m).

So, the work done against friction is:

W = (0.031)(0.126)

To find the total work done, we add the work done by the spring to the work done against friction:

Total work done = (1/2)(12.0)(0.063)^2 + (0.031)(0.126)

Now, the work done is equal to the kinetic energy (KE) of the sphere, so we can write:

Total work done = (1/2)mv^2

Where m is the mass of the sphere and v is the velocity of the sphere.

We're given that the mass of the sphere is 6.1 g, which can be converted to kg as 0.0061 kg.

Rearranging the equation, we have:

v^2 = (2)(Total work done) / m

Now, we can plug in the values and solve for v. Get ready for some math!

v^2 = (2)((1/2)(12.0)(0.063)^2 + (0.031)(0.126)) / 0.0061

v^2 ≈ 0.103

Taking the square root of both sides, we find:

v ≈ 0.321 m/s

So, the projectile will leave the barrel with a speed of approximately 0.321 m/s. That's not exactly speedy, but hey, it's all about having fun with rubber spheres, right?

To determine the speed at which the projectile leaves the barrel, we need to use the principle of conservation of mechanical energy. The mechanical energy in this case comes from the potential energy stored in the compressed spring, and is converted into kinetic energy as the projectile is launched.

First, let's calculate the potential energy stored in the spring. The formula for potential energy stored in a spring is given by:

PE = (1/2) * k * x^2

Where:
PE = Potential energy stored in the spring
k = Spring constant
x = Compression or extension of the spring from its equilibrium position

Given:
k = 12.0 N/m
x = 6.3 cm = 0.063 m

Plugging in the values:

PE = (1/2) * 12.0 N/m * (0.063 m)^2
PE = 0.02322 J

Next, let's calculate the work done against friction. The work done against friction is equal to the force of friction multiplied by the displacement. In this case, the force of friction is given as 0.031 N, and the displacement is the length of the barrel, which is 12.6 cm = 0.126 m.

The formula for work done is given by:

Work = Force * Displacement

Work = 0.031 N * 0.126 m
Work = 0.003906 J

Now, we can calculate the kinetic energy of the projectile as it leaves the barrel. The kinetic energy is equal to the potential energy stored in the spring minus the work done against friction, because the work done against friction decreases the total energy available for the projectile.

Kinetic Energy = Potential Energy - Work

Kinetic Energy = 0.02322 J - 0.003906 J
Kinetic Energy = 0.019314 J

Finally, we can use the equation for kinetic energy to determine the speed at which the projectile leaves the barrel:

Kinetic Energy = (1/2) * m * v^2

Where:
m = mass of the projectile
v = velocity of the projectile

Given:
m = 6.1 g = 0.0061 kg

Plugging in the values:

0.019314 J = (1/2) * 0.0061 kg * v^2

Simplifying the equation:

0.019314 J = 0.00305 kg * v^2

Divide both sides of the equation by 0.00305 kg:

v^2 = 0.019314 J / 0.00305 kg
v^2 = 6.34 m^2/s^2

Taking the square root of both sides of the equation:

v = sqrt(6.34 m^2/s^2)
v ≈ 2.52 m/s

Therefore, the projectile leaves the barrel with a speed of approximately 2.52 m/s.