A stone of mass 20g is released from a catapult whose rubber has been arched through 4cm. If the force constant of the rubber is 200N/M. Calculate the velocity with which the stone leaves the catapult.
vf^2=vi^2+2a*d but a = force/mass so
vf^2=2*(200/.020) * .04
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I'm sorry, I don't understand what you mean. Can you please clarify your question or provide more information?
To calculate the velocity with which the stone leaves the catapult, we can use the principle of conservation of energy.
Step 1: Convert the mass of the stone into kilograms.
Given: Mass of the stone = 20 g
1 kg = 1000 g
So, 20 g = 20/1000 kg = 0.02 kg
Step 2: Calculate the potential energy stored in the rubber band.
Potential energy (PE) = Force (F) × Displacement (d)
Given: Force constant (k) = 200 N/m
Displacement (d) = 4 cm = 4/100 m
PE = (1/2)k(d^2)
PE = (1/2) × 200 × (4/100)^2
PE = 0.04 J
Step 3: Calculate the kinetic energy of the stone when it is released.
Since energy is conserved, the potential energy stored in the rubber band is converted into kinetic energy when the stone is released.
Kinetic energy (KE) = Potential energy (PE)
KE = 0.04 J
Step 4: Calculate the velocity of the stone.
Kinetic energy (KE) = (1/2) × Mass (m) × Velocity (v)^2
0.04 J = (1/2) × 0.02 kg × v^2
0.04 J = 0.01 kg × v^2
v^2 = (0.04 J)/(0.01 kg)
v^2 = 4 m^2/s^2
v = √(4 m^2/s^2)
v = 2 m/s
Therefore, the stone leaves the catapult with a velocity of 2 m/s.