You shoot a 57 gram dart vertically upwards from a catapult with a spring constant of 379 N/m. The catapult is initially stretched from the equilibrium point by 30 cm. What is the height above the starting point reached by the dart? Ignore air resistance and give your answer to 2 s.f.
To find the height above the starting point reached by the dart, we can use the principles of conservation of mechanical energy.
Since there is no air resistance, the total mechanical energy of the dart is conserved throughout its flight. The mechanical energy is the sum of the potential energy and the kinetic energy:
Mechanical energy (E) = Potential energy (PE) + Kinetic energy (KE)
At the highest point of the dart's trajectory, all of its initial kinetic energy is converted into potential energy. Thus, we can equate the initial mechanical energy to the potential energy at the highest point:
E_initial = PE_highest point
The initial mechanical energy is given by the equation:
E_initial = 1/2kx^2
where k is the spring constant and x is the initial displacement from the equilibrium point.
In this case, the spring constant (k) is given as 379 N/m, and the initial displacement (x) is 30 cm (or 0.3 m). Plugging these values into the equation, we find:
E_initial = 1/2(379 N/m)(0.3 m)^2
= 20.385 J (rounded to three decimal places)
Since the potential energy at the highest point is equal to the initial mechanical energy, we have:
PE_highest point = 20.385 J
The potential energy (PE) of an object is equal to its mass (m) multiplied by the acceleration due to gravity (g) multiplied by its height (h):
PE = mgh
Rearranging the equation, we can solve for h:
h = PE / (mg)
Substituting the given values, we have:
h = 20.385 J / (0.057 kg)(9.8 m/s^2)
≈ 36.28 m (rounded to two significant figures)
Therefore, the height above the starting point reached by the dart is approximately 36.28 meters.
To find the height above the starting point reached by the dart, we can use the concept of potential energy and Hooke's law.
First, let's calculate the potential energy stored in the spring when it is stretched by 30 cm.
Potential energy (PE) = 1/2 * k * x^2
Where:
k = spring constant = 379 N/m
x = displacement = 0.30 m
PE = 1/2 * 379 N/m * (0.30 m)^2
PE = 1/2 * 379 N/m * 0.09 m^2
PE = 1/2 * 34.11 N * m
PE = 17.06 J (Joules)
Now, let's use the principle of conservation of energy to find the maximum potential energy of the dart at its highest point (height).
The potential energy at the highest point is converted into gravitational potential energy.
Potential Energy (PE) = m * g * h
Where:
m = mass of the dart = 57 g = 57 * 10^-3 kg = 0.057 kg
g = acceleration due to gravity = 9.8 m/s^2
h = height
Using the conservation of energy, we can equate the potential energy stored in the spring to the potential energy of the dart at its highest point:
PE (spring) = PE (dart)
17.06 J = 0.057 kg * 9.8 m/s^2 * h
Solving for h:
h = 17.06 J / (0.057 kg * 9.8 m/s^2)
h ≈ 30.19 m
Therefore, the height above the starting point reached by the dart is approximately 30.19 meters (to 2 significant figures).