Jushiro wrote:
10) 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?
Editing....
Somethings/Equations that may help you....
An ideal spring will follow Hooke's Law:
F = -k x
The work done (and therefore the stored potential energy) will then be
U = (0.5) * k * x^2
Your potential energy of the spring is (0.5) * (8.0 N/m) * (1 m / 100 cm) * (5 cm) * (5 cm). This becomes 1 N / cm.
U = 1 N / m
Now that you have solved for the potential energy of the spring, you can begin working on the velocity of the object after it exists the barrel of the gun.
Some more useful equations...
F = m * a
Total Mechanical Energy (TME) = Potential Energy (PE) + Kinetic Energy (KE)
Potential energy is m*g*h where m is the mass of the object, g is the gravitational acceleration of the object, and h is the height from the ground.
Potential spring energy is U = (0.5) * k * x^2 (shown earlier)
Kinetic energy is 1/2 * m * v^2
Going back to the TME equation.... At time equal to zero (time after compression of the spring)... the system has 1 N/cm total energy because
TME = KE + PE
TME = 0 + 1 N/cm
Therefore TME is 1 N/cm
After time equal to zero (time right upon release of the spring)... the system still has 1 N/cm total energy due to potential energy of the spring. All this energy has now been converted to Kinetic Energy.
0.1 N/cm = KE = 1/2 * m * v^2 (Original energy equation)
0.1 N/cm = 1/2 * m * v^2 (Original energy equation prettier version from above)
0.1 N/cm = 1/2 * (5.3g) * v^2 (Original energy equation subbing mass of object)
0.2 N/cm = (5.3g) * v^2 (Multiplying both sides by 2)
0.2 N/cm / 5.3g = (5.3g) * v^2 (Multiplying both sides by 2)
Conversion time....
One Newton is equivalent to 1 (kg * m) / s^2
After conversion time you get
0.3774 m^2 / s^2 = v^2
Solving for velocity upon spring release you get
0.6143 m/s as the velocity of this object upon spring release
(Just a side note... this is for a frictionless surface (Not gonna solve your problem for you but will give you an equation to get started with))
The tricky part is this
|--- spring compressed by 5 cm ---|--- barrel of gun is 15 cm ---|
^--- velocity is 0.6143 m/s as the velocity of this object upon spring release
|--- spring compressed by 5 cm ---|--- barrel of gun is 15 cm ---|
the question is what is the velocity of the object as it exits here -^
Now you have to solve for the velocity of the object in the distance of the barrel exit from the release of the spring so you have 5+15 cm to cover so 20 total centimeters to exit of barrel.
You also have the resistance against the bullet.
Useful Equations to solve for the velocity at the end of the barrel
TME = KE + PE equation
X is now 20 centimeters
Fr = (mu) * N = 0.032 N (Force Resistance Equation)
Meh, stuck with something at work. Will see if I can help a bit more later.
Use this equation to help figure your answer out... you now have all the information at the start to figure out what you need at the end.
Information for the start
TME = KE + PE equation
Information for the end (Include the force resistance here)
TME = KE + PE equation
