A stone of mass 20g is released from a catapult whose rubber is stretched through 5cm. If the force constant of the rubber is 200Nm\(^{-1}\), calculate the speed with which the stone leaves the catapult.
To solve this problem, we need to use the principle of conservation of energy, which states that the initial energy of the system is equal to the final energy of the system. Initially, the rubber band has potential energy stored in it due to its stretching, which is then transferred to the stone when it is released. This potential energy is converted into kinetic energy as the stone is propelled forward.
The potential energy stored in the rubber band can be calculated using the formula:
PE = 0.5 * k * x^2
where PE is the potential energy, k is the force constant of the rubber band, and x is the distance the rubber band is stretched.
In this case, PE = 0.5 * 200 N/m * (0.05 m)^2 = 0.5 J
Since the potential energy is converted to kinetic energy when the stone is released, we can equate the two and solve for the velocity of the stone:
PE = KE
0.5 * m * v^2 = 0.5 J
where m is the mass of the stone and v is its velocity.
Rearranging this equation, we get:
v = sqrt(2 * PE / m)
Substituting the values we have, we get:
v = sqrt(2 * 0.5 J / 0.02 kg) = 5 m/s
Therefore, the speed with which the stone leaves the catapult is 5 m/s.