a spring driven dark gun propels a .01kg dart. it has a spring constant of 784 N/m and the spring is compressed a distance of .05 m. with what speed will the dart leave the gun, assuming the spring has negligible mass?
To find the speed at which the dart leaves the gun, you can use the principle of conservation of energy. The potential energy stored in the spring when it is compressed will be converted into kinetic energy when the dart is released.
Using the elastic potential energy equation:
Elastic Potential Energy (PE) = (1/2) * k * x^2
where k is the spring constant and x is the compression distance.
The potential energy stored in the spring is given by:
PE = (1/2) * 784 N/m * (0.05 m)^2
= 0.98 J
The potential energy is equal to the kinetic energy of the dart when it is released, so:
KE = 0.98 J
The kinetic energy equation is:
KE = (1/2) * m * v^2
where m is the mass of the dart and v is its velocity.
Substituting in the given values:
0.98 J = (1/2) * 0.01 kg * v^2
Simplifying:
0.98 J = 0.005 kg * v^2
Dividing both sides by 0.005 kg:
v^2 = 196
Taking the square root of both sides:
v ≈ 14.0 m/s
Therefore, the dart will leave the gun with a speed of approximately 14.0 m/s.
To find the speed at which the dart leaves the gun, we need to apply the principle of conservation of mechanical energy. This principle states that the total mechanical energy of an object remains constant if no external forces are acting upon it.
In this case, the spring is the only source of mechanical energy, and it is initially fully compressed, converting potential energy into kinetic energy when released.
First, let's find the potential energy stored in the compressed spring:
Potential Energy (PE) = 1/2 * k * x^2
Where:
k = spring constant = 784 N/m
x = compression distance = 0.05 m
PE = 1/2 * 784 * (0.05)^2 = 0.98 J
Next, we equate the potential energy to the kinetic energy of the dart when it leaves the gun:
Kinetic Energy (KE) = 1/2 * m * v^2
Where:
m = mass of the dart = 0.01 kg (given)
v = velocity of the dart (to be determined)
Setting PE equal to KE:
0.98 J = 1/2 * 0.01 kg * v^2
Rearranging the equation to solve for v:
v^2 = (2 * PE) / m
v^2 = (2 * 0.98 J) / 0.01 kg
v^2 = 196 J/kg
v = √(196 J/kg)
Now, calculating the speed of the dart:
v ≈ √(196 J/kg) ≈ 14 m/s
Therefore, the dart will leave the gun at a speed of approximately 14 m/s.