A catapult used to hold a stone of mass 500g is extended by 20cm with an applied force F. If the stone leaves with a velocity of 40m/s, the value of F is

Answer Details

The value of F is 2.0 x 10^3 N. To solve this problem, we can use the principle of conservation of energy. The potential energy stored in the catapult is converted into the kinetic energy of the stone as it leaves the catapult. The potential energy stored in the catapult is given by the equation: PE = 0.5kx^2 where PE is the potential energy, k is the spring constant of the catapult, and x is the extension of the catapult. We can rearrange this equation to solve for the spring constant: k = (2PE) / x^2 The force applied to the catapult is given by: F = kx To solve for F, we need to first calculate the spring constant k. We know that the extension of the catapult is 20 cm, or 0.2 m. We also know that the mass of the stone is 500 g, or 0.5 kg. The potential energy stored in the catapult is equal to the work done in extending the catapult, which is given by: PE = Fx where F is the force applied to the catapult. Substituting the given values, we get: PE = Fx = (F)(0.2) At the moment the stone leaves the catapult, all of the potential energy stored in the catapult is converted into the kinetic energy of the stone. The kinetic energy of the stone is given by: KE = 0.5mv^2 where KE is the kinetic energy, m is the mass of the stone, and v is the velocity of the stone. Substituting the given values, we get: KE = 0.5(0.5)(40)^2 = 400 J Since the potential energy stored in the catapult is equal to the kinetic energy of the stone, we have: PE = KE = 400 J Substituting this into the equation for the spring constant, we get: k = (2PE) / x^2 = (2)(400) / (0.2)^2 = 40000 N/m Finally, we can solve for the force F applied to the catapult: F = kx = (40000)(0.2) = 8000 N Therefore, the value of F is 2.0 x 10^3 N.