A toy gun launches a small plastic sphere using a spring. The spring has a spring constanct of 20 J/M^2 and the plastic sphere has a mass of 25 grams.
Before launch
x initial is -10cm
velocity initial is 0 m/s
After launch
x final is o cm
Apply the law of conservation of energy to find the speed of the sphere just as it leaves the gun.
1/2 k x^2=1/2 m v^2
x=.1meter solve for v.
To find the speed of the sphere just as it leaves the gun, we can apply the law of conservation of energy, which states that energy cannot be created or destroyed, only transferred or transformed from one form to another.
In this case, the initial energy of the system is stored in the spring potential energy before launch. The final energy of the system will be entirely kinetic energy of the sphere just as it leaves the gun, assuming there is no other energy loss (such as due to air resistance).
The equation for spring potential energy is given by:
PE = (1/2)kx^2
Where PE is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position (in this case, the compressed position of the spring).
Given that the spring constant is 20 J/m^2 and the initial displacement is -10 cm (-0.1 m), we can substitute these values into the equation:
PE = (1/2)(20 J/m^2)(-0.1 m)^2
PE = (1/2)(20 J/m^2)(0.01 m^2)
PE = 0.1 J
This is the initial potential energy stored in the spring.
Now, since energy is conserved, the initial potential energy must be equal to the final kinetic energy of the sphere just as it leaves the gun. The equation for kinetic energy is:
KE = (1/2)mv^2
Where KE is the kinetic energy, m is the mass of the sphere, and v is the velocity.
Given that the mass of the sphere is 25 grams (0.025 kg) and the final displacement is 0 cm (0 m), we can substitute these values into the equation:
KE = (1/2)(0.025 kg)v^2
Since the final displacement is 0 cm, the velocity will be the only unknown, denoted as v.
So, we have the equation:
0.1 J = (1/2)(0.025 kg)v^2
To solve for v, divide both sides of the equation by (1/2)(0.025 kg):
0.1 J / [(1/2)(0.025 kg)] = v^2
Simplifying:
4 J/kg = v^2
Taking the square root of both sides:
√(4 J/kg) = √(v^2)
2 (J/kg)^(1/2) = v
Therefore, the speed of the sphere just as it leaves the gun is 2 (J/kg)^(1/2).