Calculating Potential Energy - High School Physics

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Question

A skier is at the top of a hill. At the bottom of the hill, she has a velocity of $\small 12\frac{m}{s}$ . How tall was the hill?

Answer

At the top of the hill the skier has purely potential energy. At the bottom, she has purely kinetic energy.

We can solve by understanding the conservation of energy. The skier's energy at the top of the hill will be equal to her energy at the bottom of the hill.

$PE_{top}=KE_{bottom}$

Using the equations for potential and kinetic energy, we can solve for the height of the hill.

$mgh=\frac{1}{2}mv^2$

The masses cancel, and we can plug in our final velocity and gravitational acceleration.

$gh=\frac{1}{2}v^2$

$(-9.8\frac{m}{s^2})h=\frac{1}{2}(12\frac{m}{s})^2$

$(-9.8\frac{m}{s^2})h=\frac{1}{2}144\frac{m^2}{s^2}$

$(-9.8\frac{m}{s^2})h=72\frac{m^2}{s^2}$

$h=\frac{72\frac{m^2}{s^2}}{-9.8\frac{m}{s^2}}$

This formula solves for the change in height. The negative sign implies she travelled in a downward direction. Because the question is asking how tall the hill is, we use an absolute value.

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Question

An astronaut is on a new planet. She discovers that if she drops a space rock from above the ground, it has a final velocity of $3\frac{m}{s}$ just before it strikes the planet surface. What is the acceleration due to gravity on the planet?

Answer

We can use conservation of energy to solve. The potential energy when the astronaut is holding the rock will be equal to the kinetic energy just before it strikes the surface.

$mgh=\frac{1}{2}mv^2$

Now, we need to solve for $\small g$ , the gravity on the new planet. The masses will cancel out.

$gh=\frac{1}{2}v^2$

Plug in the given values and solve.

$g(-10m)=\frac{1}{2}(3\frac{m}{s})^2$

$g(-10m)=\frac{1}{2}(9\frac{m^2}{s^2})$

$g(-10m)=(4.5\frac{m^2}{s^2})$

$g=\frac{4.5\frac{m^2}{s^2}}{-10m}$

$g=-0.45\frac{m}{s^2}$

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Question

A ball is about to roll off the edge of a tall table. What is its current potential energy?

$\small g=-9.8\frac{m}{s^2}$

Answer

The equation for potential energy is . We are given the mass of the ball, the height of the table, and the acceleration of gravity in the question. The distance the ball travels is in the downward direction, making it negative.

Plug in the values, and solve for the potential energy.

$PE=(0.5kg)(-9.8\frac{m}{s^2})(-1.5m)$

The units for energy are Joules.

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Question

A pendulum with string length is dropped from rest. If the mass at the end of the pendulum is , what is its initial potential energy?

$\small g=-9.8\frac{m}{s^2}$

Answer

Potential energy can be found using the equation . For the pendulum, the height is going to be the length of the string.

Remember, your height is your change in distance. In this case the ball will go down , so the height will be negative since the ball travels downward.

$PE= (2.03kg)(-9.8\frac{m}{s^2})(-1.2m)$

The units for energy are Joules.

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Question

A ball rolls up a hill. If the ball is initially travelling with a velocity of $3.12\frac{m}{s}$ , how high up the hill does it roll?

$\small g=-9.8\frac{m}{s^2}$

Answer

Use the conservation of energy equation to solve for the potential energy at the top of the hill.

$PE_{top}=KE_{bottom}$

$mgh=\frac{1}{2}mv^2$

Plug in the values given to you and solve for the final height.

$(0.5kg)(9.8\frac{m}{s^2})(h)=\frac{1}{2}(0.5kg)(3.12\frac{m}{s})^2$

$4.9\frac{kg\cdot m}{s^2}h=2.4336\frac{kg\cdot m^2}{s^2}$

$h=\frac{2.4336\frac{kg\cdot m^2}{s^2}}{4.9\frac{kg\cdot m}{s^2}}$

$h=0.497m\approx0.5m$

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Question

Rob throws a ball vertically in the air with an initial velocity of $3.5\frac{m}{s}$ . What is the maximum height of the ball?

$\small g=-9.8\frac{m}{s^2}$

Answer

The maximum height will be when the ball has only potential energy, and no kinetic energy. Initially, the ball has only kinetic energy and no potential energy. We can set these values equal to each other due to conservation of energy.

$PE_{top}=KE_{bottom}$

$mgh=\frac{1}{2}mv^2$

The masses will cancel out.

$gh=\frac{1}{2}v^2$

Plug in the values that were given and solve for the height.

$(9.8\frac{m}{s^2})h=\frac{1}{2}(3.5\frac{m}{s})^2$

$(9.8\frac{m}{s^2})h=\frac{1}{2}(12.25\frac{m^2}{s^2})$

$(9.8\frac{m}{s^2})h=6.125\frac{m^2}{s^2}$

$h=\frac{6.125\frac{m^2}{s^2}}{9.8\frac{m}{s^2}}$

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Question

Laurence throws a rock off the edge of a tall building at an angle of $20^{\circ}$ from the horizontal with an initial speed of $31\frac{m}{s}$ .

$g=-9.8\frac{m}{s^2}$ .

What is the potential energy of the rock at the moment it is released?

Answer

The formula for gravitational potential energy is:

We can solve for this value using the given mass of the rock, acceleration of gravity, and initial height.

$PE=(0.5kg)(9.8\frac{m}{s^2})(12m)$

This value is independent of the kinetic energy of the rock, and is not dependent on initial velocity.

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Question

Sam throws a rock off the edge of a tall building at an angle of $35^{\circ}$ from the horizontal. The rock has an initial speed of $30\frac{m}{s}$ .

$\small g=-9.8\frac{m}{s^2}$

What is the gravitational potential energy of the rock as soon as it leaves Sam's hand?

Answer

The formula for gravitational potential energy is . The only relevant variables are mass, gravity, and height. All of the information about velocity and angle are not needed to solve for the initial potential energy.

Plug in the given values for mass and height, and solve using the equation. Keep in mind that the height will be negative because the rock travels downward.

$PE=(0.5kg)(-9.8\frac{m}{s^2})(-10m)$

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Question

A ball sits at rest on a table above the ground. What is the potential energy of the ball?

$\small g=-9.8\frac{m}{s^2}$

Answer

The only potential energy this ball can have is gravitational potential energy. The formula for gravitational potential energy is .

We are given the height and mass of the ball. Using the given values, we can solve for the potential energy.

Keep in mind that the displacement will be negative because the ball is traveling in the downward direction.

$PE=(1.11kg)(-9.8\frac{m}{s^2})(-1.72m)$

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Question

Two balls, one with mass and one with mass , are dropped from above the ground. What is the potential energy of the ball right before it starts to fall?

$\small g=-9.8\frac{m}{s^2}$

Answer

The equation for potential energy is .

Since we know the mass, height, and acceleration from gravity, we can simply multiply to find the potential energy.

$PE=(2kg)(-9.8\frac{m}{s^2})(-30m)$

Note that we plugged in for because the ball will be moving downward; the change in height is negative as the ball drops.

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Question

Two balls, one with mass and one with mass , are dropped from above the ground. What is the potential energy of the ball right before it starts to fall?

$\small g=-9.8\frac{m}{s^2}$

Answer

The equation for potential energy is .

Since we know the mass, height, and acceleration from gravity, we can simply multiply to find the potential energy.

$PE=(4kg)(-9.8\frac{m}{s^2})(-30m)$

Note that we plugged in for because the ball will be moving downward; the change in height is negative as the ball drops.

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Question

A book falls off the top of a bookshelf. What is its potential energy right before it falls?

$\small g=-9.8\frac{m}{s^2}$

Answer

The formula for potential energy is .

Given the values for the mass, height, and gravity, we can solve using multiplication. Note that the height is negative because the book falls in the downward direction.

$PE=2kg-9.8\frac{m}{s^2}-2.3m$

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Question

A book falls off the top of a bookshelf. What is its potential energy right before it falls?

$\small g=-9.8\frac{m}{s^2}$

Answer

The formula for potential energy is .

Given the values for the mass, height, and gravity, we can solve using multiplication. Note that the height is negative because the book falls in the downward direction.

$PE=1.12kg-9.8\frac{m}{s^2}-2.03m$

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Question

A book falls off the top of a bookshelf. What is its potential energy right before it falls?

$\small g=-9.8\frac{m}{s^2}$

Answer

The formula for potential energy is .

Given the values for the mass, height, and gravity, we can solve using multiplication. Note that the height is negative because the book falls in the downward direction.

$PE=3.23kg-9.8\frac{m}{s^2}-3.01m$

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Question

A box is placed on top of a $\small 3m$ tall spring resting on the ground. The box has a weight of $\small 10N$ , and compresses the spring $\small 1m$ . What is the spring constant?

Answer

We need two formulas for this problem: the gravitational potential energy and the spring potential energy.

$PE_s = \frac{1}{2}kx^2$

We know the weight of the box and the change in height when it is placed on the spring. These values allow us to calculate the change in potential energy. Due to conservation of energy, the total energy must remain constant. Initially, the box only has gravitational potential energy. In its final position, it has both gravitational and spring potential energy.

$PE_{g1}=PE_{g2}+PE_s$

$mgh_1=mgh_2+\frac{1}{2}kx^2$

We know the initial height and final height of the spring. This also gives us the value of $\small x$ , the displacement of the spring.

$mg(3m)=mg(2m)+\frac{1}{2}k(1m)^2$

We also know the weight of the box: $\small mg=10N$

$(10N)(3m)=(10N)(2m)+\frac{1}{2}k(1m)^2$

Now we can solve for the spring constant, $\small k$ .

$30J=20J+\frac{1}{2}k(1m^2)$

$10J=(\frac{1}{2}m^2)k$

$20\frac{N}{m}=k$

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Question

What is the potential energy of a ball held above the ground?

$g=-9.8\frac{m}{s^2}$

Answer

The formula for gravitational potential energy is:

We are given the mass of the ball, height, and acceleration from gravity. Remember, since the displacement is downward it must be negative.

$PE=0.5kg-9.8\frac{m}{s^2}-3m$

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Question

Calculate the potential energy of a $\small 500g$ branch attached to a tree at a point $\small 30m$ above the ground.

Answer

Potential energy due to gravity is given by the equation:

We are given the mass of the branch and its height. Gravity is constant. Using these values, we can solve for the potential energy.

First, convert the mass of the branch to kilograms.

$500g*\frac{1kg}{1000g}=0.5kg$

Then, use the equation to find the energy.

$PE=(0.5kg)(9.8\frac{m}{s^2})(30m)$

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Question

A man stands at the top of a tall building. He holds a rock over the edge. What is the potential gravitational energy of the rock?

Answer

Potential gravitational energy is given by the equation:

We are told the height of the rock and its mass. Using the constant acceleration due to gravity, we can solve for the gravitational potential energy.

$PE=(3.2kg)(9.8\frac{m}{s^2})(21m)$

$PE=658.56\ \frac{kg\cdot m^2}{s^2}=658.56J$

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