Calculate the potential energy stored in an object due to its position. Enter mass, height, and gravity for instant results, useful in physics and engineering.
Calculate the potential energy of a 15 kg object located 25 meters above the ground. Assume \(g = 9.81 \, \text{m/s}^2\).
Solution:
The formula for gravitational potential energy is:
\(PE = mgh\)
Substituting the given values:
\(PE = 15 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 25 \, \text{m}\)
So, the potential energy is:
\(PE = 3686.25 \, \text{J}\)
The Earth-Moon distance is 384,400 km. If the mass of the Earth is \(5.972 \times 10^{24} \, \text{kg}\) and the mass of the Moon is \(7.347 \times 10^{22} \, \text{kg}\), calculate the gravitational potential energy of the Earth-Moon system. Use \(G = 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2\).
Solution:
The formula for gravitational potential energy between two masses is:
\(PE = -\frac{G m_1 m_2}{r}\)
Substituting the given values:
\(PE = -\frac{6.674 \times 10^{-11} \times 5.972 \times 10^{24} \times 7.347 \times 10^{22}}{384,400,000}\)
So, the potential energy is approximately:
\(PE = -1.982 \times 10^{29} \, \text{J}\)
Calculate the elastic potential energy stored in a spring with spring constant \(k = 250 \, \text{N/m}\) when it is compressed by 0.2 m.
Solution:
The formula for elastic potential energy is:
\(PE = \frac{1}{2} k x^2\)
Substituting the given values:
\(PE = \frac{1}{2} \times 250 \, \text{N/m} \times (0.2 \, \text{m})^2\)
So, the potential energy is:
\(PE = 5 \, \text{J}\)
Calculate the potential energy of a pendulum of length 2 m and mass 0.5 kg when it is raised 1.5 m above its lowest point. Assume \(g = 9.81 \, \text{m/s}^2\).
Solution:
The formula for gravitational potential energy is:
\(PE = mgh\)
Substituting the given values:
\(PE = 0.5 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 1.5 \, \text{m}\)
So, the potential energy is:
\(PE = 7.3575 \, \text{J}\)
A roller coaster car of mass 400 kg is at the top of a 50-meter hill. Calculate its potential energy at the top and its kinetic energy at the bottom of the hill (assuming no friction). Use \(g = 9.81 \, \text{m/s}^2\).
Solution:
The potential energy at the top is:
\(PE_{\text{top}} = mgh\)
\(PE_{\text{top}} = 400 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 50 \, \text{m}\)
At the bottom of the hill, all the potential energy is converted to kinetic energy (ignoring friction), so:
\(KE_{\text{bottom}} = 400 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 50 \, \text{m} = 196,200 \, \text{J}\)
An object has a gravitational potential energy of 300 J. If its mass is 10 kg, calculate the height from which it is suspended. Assume \(g = 9.81 \, \text{m/s}^2\).
Solution:
The formula for gravitational potential energy is:
\(PE = mgh\)
Rearranging to solve for height:
\(h = \frac{PE}{mg}\)
Substituting the given values:
\(h = \frac{300 \, \text{J}}{10 \, \text{kg} \times 9.81 \, \text{m/s}^2}\)
So, the height is:
\(h = 3.06 \, \text{m}\)
Calculate the velocity of a 0.3 kg object that is released from a spring compressed by 0.5 m, with a spring constant of 150 N/m. Assume all potential energy is converted into kinetic energy.
Solution:
The elastic potential energy in the spring is:
\(PE_{\text{spring}} = \frac{1}{2} k x^2\)
\(PE_{\text{spring}} = \frac{1}{2} \times 150 \, \text{N/m} \times (0.5 \, \text{m})^2\)
The potential energy is converted to kinetic energy:
\(KE = \frac{1}{2} mv^2\)
Equating the two energies and solving for \(v\):
\(\frac{1}{2} mv^2 = \frac{1}{2} k x^2\)
\(v = \sqrt{\frac{k x^2}{m}}\)
Substituting the given values:
\(v = \sqrt{\frac{150 \times (0.5)^2}{0.3}}\)
So, the velocity is:
\(v \approx 5.48 \, \text{m/s}\)
How much work is required to lift a 25 kg object 10 meters vertically? Assume \(g = 9.81 \, \text{m/s}^2\).
Solution:
The work done is equal to the change in potential energy:
\(W = mgh\)
Substituting the given values:
\(W = 25 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 10 \, \text{m}\)
So, the work done is:
\(W = 2450 \, \text{J}\)
Calculate the potential energy of a satellite of mass 500 kg orbiting at a distance of 10,000 km from Earth's center. Use the gravitational constant \(G = 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2\) and Earth's mass \(M = 5.972 \times 10^{24} \, \text{kg}\).
Solution:
The formula for gravitational potential energy is:
\(PE = -\frac{GMm}{r}\)
Substituting the given values:
\(PE = -\frac{6.674 \times 10^{-11} \times 5.972 \times 10^{24} \times 500}{10,000,000}\)
So, the potential energy is approximately:
\(PE = -1.99 \times 10^9 \, \text{J}\)
A ball is dropped from a height of 30 m. What is its speed just before hitting the ground, assuming no air resistance and mass of 2 kg?
Solution:
The potential energy at the height is:
\(PE = mgh\)
The ball's speed just before hitting the ground will be found by equating the potential energy to kinetic energy:
\(KE = \frac{1}{2} mv^2\)
Equating the two energies:
\(mgh = \frac{1}{2} mv^2\)
Solving for \(v\):
\(v = \sqrt{2gh}\)
Substituting the values:
\(v = \sqrt{2 \times 9.81 \times 30}\)
So, the velocity is:
\(v \approx 24.26 \, \text{m/s}\)