A spring with a force constant of 5000N/m and a rest length of 3m is used in a catapult. When compressed to 1m, it is used to launch a 50kg rock. However, there is an error in the release mechanism, so the rock gets launched almost straight up. How high does it go, and how fast is it going when it hits the ground?
To solve this problem, we can apply the principles of energy conservation. At the moment the rock is released, the energy stored in the compressed spring is completely converted into the kinetic energy of the rock. When the rock reaches its highest point, all of its kinetic energy will be converted into gravitational potential energy. Thus, we can equate the initial potential energy of the spring to the final potential energy of the rock at the highest point:
Initial potential energy = Final potential energy
Let's calculate the initial and final potential energies step by step.
Step 1: Calculate the potential energy stored in the compressed spring.
The formula for potential energy of a spring is given by:
Potential energy = (1/2) * k * x^2
Where:
k = force constant of the spring (5000 N/m)
x = compression of the spring (1 m)
Potential energy = (1/2) * 5000 N/m * (1 m)^2
Step 2: Calculate the initial kinetic energy of the rock.
The formula for kinetic energy is given by:
Kinetic energy = (1/2) * m * v^2
Where:
m = mass of the rock (50 kg)
v = velocity of the rock when released
Since the spring's potential energy is completely converted into kinetic energy, we can set the potential energy equal to the kinetic energy:
(1/2) * 5000 N/m * (1 m)^2 = (1/2) * 50 kg * v^2
Step 3: Solve for the velocity of the rock when released.
Rearrange the equation and solve for v:
v^2 = (5000 N/m * (1 m)^2 * 2) / (50 kg)
v^2 = 2000 m^2/s^2
v ≈ 44.72 m/s
Therefore, the velocity of the rock when released is approximately 44.72 m/s.
Step 4: Calculate the highest point reached by the rock.
At the highest point, the gravitational potential energy is maximum, while the kinetic energy is zero. The gravitational potential energy is given by:
Gravitational potential energy = m * g * h
Where:
m = mass of the rock (50 kg)
g = acceleration due to gravity (approximated to 9.8 m/s^2)
h = height above the ground at the highest point
Since the kinetic energy is zero at the highest point, we can equate the initial potential energy to the gravitational potential energy:
(1/2) * 5000 N/m * (1 m)^2 = 50 kg * 9.8 m/s^2 * h
Solving for h:
h = (5000 N/m * (1 m)^2) / (50 kg * 9.8 m/s^2)
h ≈ 10.20 m
Therefore, the rock reaches a height of approximately 10.20 meters.
Step 5: Calculate the speed of the rock when it hits the ground.
We can use kinematic equations to find the speed of the rock when it hits the ground. The equation that relates the initial velocity (44.72 m/s), height (10.20 m), and final velocity (v' = 0 m/s) is given by:
v'^2 = v^2 + 2 * g * h
Solve for v':
v'^2 = 44.72 m/s^2 + 2 * 9.8 m/s^2 * 10.20 m
v'^2 = 44.72 m^2/s^2 + 199.92 m^2/s^2
v' ≈ 18.96 m/s
Therefore, the rock will be moving at approximately 18.96 m/s when it hits the ground.