Calculate Potential Energy, Mass, Height, or Gravity with different units.
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:
\(v \approx 24.26 \, \text{m/s}\)