A slingshot is pulled back 55cm down to the ground with a rock in its pouch. The slingshot has a spring constant of 862N/m. The mass of the rock is 700g. What height is the rock when its velocity is 12m/s?
To determine the height of the rock when its velocity is 12m/s, we can use the principles of mechanical energy.
First, let's determine the potential energy of the slingshot and the rock when it is pulled back. The potential energy is given by the equation:
Potential Energy = (1/2) * k * x^2
where k is the spring constant and x is the displacement of the slingshot from its equilibrium position.
In this case, the displacement is 55cm or 0.55m, and the spring constant is 862N/m. Plugging these values into the equation, we get:
Potential Energy = (1/2) * 862N/m * (0.55m)^2 = 130.355 J
Since the rock is at the highest point of its trajectory when its velocity is 12m/s, we can equate the potential energy to the kinetic energy at this point.
Kinetic Energy = (1/2) * m * v^2
where m is the mass of the rock and v is its velocity. In this case, the mass is 700g or 0.7kg, and the velocity is given as 12m/s. Plugging these values into the equation, we have:
130.355 J = (1/2) * 0.7kg * (12m/s)^2
Now, we can solve for the height h, which represents the displacement of the rock from the ground at this point.
Potential Energy = Kinetic Energy
130.355 J = (1/2) * 0.7kg * (12m/s)^2
130.355 J = 0.35kg * (12m/s)^2
130.355 J = 0.35kg * 144 m^2/s^2
130.355 J = 50.4 kg m^2/s^2
Since both sides have the same unit of energy, we can equate them:
50.4 kg m^2/s^2 = m * g * h
where g represents the acceleration due to gravity and h is the height we want to find.
Rearranging the equation to solve for h:
h = (50.4 kg m^2/s^2) / (m * g)
The mass m is 0.7kg, and the acceleration due to gravity g is approximately 9.8 m/s^2. Plugging these values into the equation, we get:
h = (50.4 kg m^2/s^2) / (0.7kg * 9.8 m/s^2) ≈ 7.8 m
Therefore, the height of the rock when its velocity is 12m/s is approximately 7.8 meters.