An amount of work equal to 3.5 J is required to compress the spring in a marble gun. If all energy is conserved, how much kinetic energy does a 25 gram marble have when it leaves the spring?
To find the kinetic energy of the marble when it leaves the spring, we need to use the principle of conservation of energy. The initial potential energy stored in the compressed spring is converted into kinetic energy as the marble is launched.
The formula to calculate the potential energy stored in a spring is given by:
Potential Energy (PE) = (1/2) * k * x^2
where k represents the spring constant and x is the distance the spring is compressed.
From the problem statement, we are given that the amount of work (W) required to compress the spring is 3.5 J. The work done on an object can also be calculated as the change in potential energy (PE) from the initial state to the final state.
So, we have:
W = PE_final - PE_initial
Since the initial potential energy is zero when the spring is uncompressed, we have:
3.5 J = PE_final - 0
Therefore, PE_final = 3.5 J.
Now, we can equate the potential energy stored in the spring to the kinetic energy of the marble:
PE_final = Kinetic Energy (KE)
Therefore, the kinetic energy of the marble when it leaves the spring is 3.5 J.
However, it's important to note that the mass of the marble is given as 25 grams. To convert the mass to kilograms, we divide by 1000:
Mass (m) = 25 grams / 1000 = 0.025 kg
The formula for kinetic energy is given by:
KE = (1/2) * m * v^2
where m is the mass of the object and v is its velocity.
Since we've already determined that the kinetic energy is 3.5 J, we can rearrange the equation to solve for velocity (v):
3.5 J = (1/2) * 0.025 kg * v^2
Now, solving for v:
v^2 = (3.5 J) / [(1/2) * 0.025 kg]
v^2 = 140 m^2/s^2
Finally, taking the square root of both sides to find v:
v = sqrt(140 m^2/s^2)
Therefore, the velocity of the marble when it leaves the spring is sqrt(140 m^2/s^2) or approximately 11.83 m/s.