Energy and Work - Physics

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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

Mike jumps off a bridge with a bungee cord (a heavy stretchable cord) tied around his ankle). He falls for before the bungee cord begins to stretch. Mike's mass is and we assume the cord obeys Hooke’s law. The constant is . If we neglect air resistance, what is the distance below the bridge Mike’s foot will be before coming to a stop. Ignore the mass of the cord and treat Mike as a particle.

Answer

We must consider several points during Mike’s jump off of the bridge. The first point is when he is at the top of the bridge when he is about to jump. The second point is the below the bridge, just when the bungee cord would begin to stretch. The third is the point at the bottom of the cord when it is fully stretched out.

To start let, us consider the first two points, when he jumps off the bridge and when he reaches below the bridge. For this first consideration, I will assume that our zero point of reference is below the bridge.

At the top of the bridge, Mike has gravitational potential energy. later, all of this potential energy has been converted to kinetic energy. According to the law of conservation of energy we can set these two things equal to each other.

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

Since mass is in both sides of the equation it can be cancelled out to leave us with

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

We can now solve for the final velocity, just before the cord stretches.

$(9.8m/s^2)(12m) = \frac{1}{2}v^2$

$117.6 = \frac{1}{2}v^2$

$\sqrt{235.2} = v$

Now let us consider two new points, the point at which the cord starts to stretch, and the point at the bottom when the entire cord is stretched out. We will consider the lowest point as our zero point of reference in this case.

At the top, Mike has kinetic energy and gravitational potential energy as he is moving and above our reference point. At the bottom all of this energy has converted to elastic potential energy. According to the law of conservation of energy we can set these two things equal to each other.

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

The cord is going to stretch the same distance that Mike starts above the ground so we can exchange our x value for h so that everything is in similar terms.

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

We can now put in our values and start to solve for h. We will use our velocity from the first part as the velocity that Mike has.

$(73kg)(9.8m/s^2)h + \frac{1}{2}(73kg)(15.34m/s)^2 = \frac{1}{2}(50N/m)h^2$

We are left with a quadratic equation. So we will need to get everything over to one side and use our quadratic formula to solve this problem.

The quadratic formula is

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-(715.4)\pm\sqrt{(-715.4)^2-4(25)(8589.02)}}{2(25)}$

The two answer we get for this is and . The reasonable answer is . This is the distance the cord will stretch.

To find the total distance below the bridge we will need to add the amount that the cord stretched to the it took to fall before the cord stretched.

Mike will stop below the bridge.

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Question

Which of the following is not an example of potential energy?

Answer

The running woman has kinetic energy as she is moving.

The candy bar has chemical potential energy.

The apple has gravitational potential energy.

The rubber band and the spring both have elastic potential energy.

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Question

A mass is put at the end of a spring with a spring constant of $180\frac{N}{m}$ . The spring is then compressed . What is the maximum velocity of the mass?

Answer

For this problem, we're going to use the law of conservation of energy. Since we're looking for max velocity, we're going to say that the of the system.

The formula for potential energy of a spring is $PE=\frac{1}{2}kx^2$

Therefore:

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

Notice that the $\frac{1}{2}$ 's cancel out.

Plug in our given values.

$180\frac{N}{m}(0.89m)^2=2kgv^2$

$180\frac{N}{m}0.7921m^2=2kgv^2$

$\frac{142.578J}{2kg}=v^2$

$71.289\frac{m^2}{s^2}=v^2$

$\sqrt{71.289\frac{m^2}{s^2}}=\sqrt{v^2}$

$8.44\frac{m}{s}=v$

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Question

Which of the following is an example of work?

Answer

Work is a force times a distance or .

Only one of these has a force times a distance: the weightlifter lifting a weight above his head.

The car moving at a constant velocity has no acceleration and therefore no force.

Typing essays does not require a force times a distance (though it sure feels like work!).

If the skier has no loss in energy, then no work was done, as work is also the measurement of the change in energy.

The man sitting in a chair has a constant velocity of zero, no acceleration, and therefore no force.

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Question

A rock is dropped from a helicopter hovering at above the ground. If the rock weighs , what is its kinetic energy right before it hits the ground?

Answer

Right before it hits the ground, the initial potential energy and the final kinetic energy will equal each other due to conservation of energy.

If we solve for initial potential, we can find final kinetic energy.

Plug in the values given. Remember that height is the change in height. Since the rock is headed downward, the height will be negative.

$(0.2kg)(-9.8\frac{m}{s^2})(-12m)=PE=KE$

Multiply and solve.

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Question

A ball is dropped from above the ground. Assuming gravity is $-9.8\frac{m}{s^2}$ , what is its final velocity?

Answer

We can use potential energy to solve. Remember, your height and your gravity need to have the same sign, as they are moving in the same direction (downward). Either make them both negative, or use an absolute value.

Using conservation of energy, we know that $PE_{top}=KE_{bottom}$ . This tells us that the potential energy at the top of the hill is all converted to kinetic energy at the bottom of the hill. We can substitute the equations for potential energy and kinetic energy.

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

The masses cancel out.

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

Plug in the values, and solve for the velocity.

$(-9.8\frac{m}{s^2})(-22m)=\frac{1}{2}v^2$

$215.6\frac{m^2}{s^2}=\frac{1}{2}v^2$

$431.2\frac{m^2}{s^2}=v^2$

$\sqrt{431.2\frac{m^2}{s^2}}=\sqrt{v^2}$

$20.77\frac{m}{s}=v$

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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 maximum velocity?

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

Answer

The maximum velocity of the pendulum will be when the object has only kinetic energy. Using conservation of energy, we can set our initial potential energy to equal our "final" kinetic energy.

$PE_{top}=KE_{bottom}$

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

Plug in the given values and solve for the velocity.

$(2.03kg)(-9.8\frac{m}{s^2})(-1.2m)=\frac{1}{2}(2.03kg)v^2$

$23.87\frac{kg\cdot m^2}{s^2}=(1.015kg)v^2$

$23.52\frac{m^2}{s^2}=v^2$

$\sqrt{23.52\frac{m^2}{s^2}}=\sqrt{v^2}$

$4.85\frac{m}{s}=v$

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Question

A ball rolls down a hill with an initial velocity of $2.45\frac{m}{s}$ . What is its maximum velocity?

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

Answer

For this problem, the ball starts with both potential and kinetic energy. The point of maximum velocity will have no potential energy. We can solve setting the initial energy and final energy equal, due to conservation of energy.

$mgh_i+\frac{1}{2}mv_i^2=mgh_f+\frac{1}{2}mv_f^2$

The masses will cancel out from all of the terms.

$gh_i+\frac{1}{2}v_i^2=gh_f+\frac{1}{2}v_f^2$

Plug in the given values and solve for the final velocity. Remember, when the ball is on the ground it has a height of zero.

$(9.8\frac{m}{s^2})(1m)+\frac{1}{2}(2.45\frac{m}{s})^2=(9.8\frac{m}{s^2})(0m)+\frac{1}{2}v_f^2$

$9.8\frac{m^2}{s^2}+\frac{1}{2}(6.0025\frac{m^2}{s^2})=0+\frac{1}{2}v_f^2$

$12.80125\frac{m^2}{s^2}=\frac{1}{2}v_f^2$

$25.6025\frac{m^2}{s^2}=v_f^2$

$\sqrt{25.6025\frac{m^2}{s^2}}=\sqrt{v_f^2}$

$5.06\frac{m}{s}=v_f$

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Question

A book falls off of a high table. If the book weighs , what will its kinetic energy be right before it hits the ground?

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

Answer

Initially, the book has only potential energy. Right before it hits the ground, all the potential energy will have converted to kinetic energy. The two values will be equal based on the conservation of energy.

If we solve for potential, we can find kinetic energy.

$(0.75kg)(9.8\frac{m}{s^2})(2.2m)=PE=KE$

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Question

A book falls off of a high table. If the book weighs , what will its final velocity be right before it hits the ground?

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

Answer

The book initially has only potential energy. Right before it hits the ground, all the potential energy will be converted to kinetic energy. We can use the law of conservation of energy to set the initial and final energies equal.

$PE_{initial}=KE_{final}$

Use the equations for potential energy and kinetic energy.

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

Now, plug in the values given to you in the problem and solve for the velocity.

$(0.75kg)(9.8\frac{m}{s^2})(2.2m)=\frac{1}{2}(0.75kg)v^2$

$\frac{16.17J}{0.375kg}=v^2$

$43.12\frac{m^2}{s^2}=v^2$

$\sqrt{43.12\frac{m^2}{s^2}}=\sqrt{v^2}$

$6.57\frac{m}{s}=v$

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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 final vertical kinetic energy right before the rock hits the ground?

Answer

The formula for kinetic energy is $KE=\frac{1}2mv^2{}$ .

We first need to find the initial velocity in the vertical direction. To find the vertical velocity we use the equation $v_{i}* \sin(\Theta)=v_{iy}$ .

We can plug in the given values for the angle and initial velocity to solve.

$30\frac{m}{s}* \sin(35^{\circ})=v_{iy}$

$30\frac{m}{s}* (0.57)=v_{iy}$

$17.21\frac{m}{s}=v_{iy}$

We know that the rock is going to travel a net distance of , as that is the distance between where the rock's initial and final positions. We now know the displacement, initial velocity, and acceleration, which will allow us to solve for the final velocity.

$v_f^2=v_1^2+2a\Delta y$

$v_f^2=(17.21\frac{m}{s})^2+2(-9.8\frac{m}{s^2})(-10m)$

$v_f^2=(296.18\frac{m^2}{s^2})+196\frac{m^2}{s^2}$

$v_f^2=492.18\frac{m^2}{s^2}$

$\sqrt{v_f^2}=\sqrt{492.18\frac{m^2}{s^2}}$

$v_{fy}=22.19\frac{m}{s}$

Now we have the mass and final velocity, allowing us to solve for the final kinetic energy.

$KE_f=\frac{1}{2}mv_f^2$

$KE_f=\frac{1}{2}(0.5kg)(22.19\frac{m}{s})^2$

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Question

A book falls off the top of a bookshelf. What is its velocity right before it hits the ground?

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

Answer

The relationship between velocity and energy is:

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

We know the mass, but we need to find the total kinetic energy.

Remember the law of conservation of energy: the total energy at the beginning equals the total energy at the end. In this case, we have only potential energy at the beginning and only kinetic energy at the end. (The initial velocity is zero, and the final height is zero).

If we can find the potential energy, we can find the kinetic energy. The formula for potential energy is .

Using our given values for the mass, height, and gravity, we can solve using multiplication. Note that the height becomes negative because the book is traveling in the downward direction.

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

The kinetic energy will also equal , due to conservation of energy.

Using this value and our given mass, we can calculate the velocity from our original kinetic energy equation.

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

$45.08J=\frac{1}{2}2kgv^2$

$45.08\frac{m^2}{s^2}=v^2$

$\sqrt{45.08\frac{m^2}{s^2}}=\sqrt{v^2}$

Since we are taking the square root, our answer can be either negative or positive. The final velocity of the book will be in the downward direction; thus, our final velocity should be negative.

$-6.71\frac{m}{s}=v$

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Question

Starting from rest a ball with a mass of 3kg experiences a constant force of 9N for 4s. Find the final kinetic energy (in joules) of the ball after 4s.

Answer

Use Newton's laws to find the acceleration of the ball. Find the final speed of the ball, and then solve for kinetic energy.

Acceleration:

$9N=(3kg)a\rightarrow a=\frac{9N}{3kg}=3\frac{m}{s^2}$

Final speed:

$V=V_{0}+at$

$V=0+\left(3\frac{m}{s^2}\right)4s=12\frac{m}{s}$

Kinetic energy:

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

$KE=\frac{1}{2}(3kg)\left(12\frac{m}{s}^2\right)=216J$

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Question

A book falls off the top of a bookshelf. What is its velocity right before it hits the ground?

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

Answer

The relationship between velocity and energy is:

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

We know the mass, but we need to find the total kinetic energy.

Remember the law of conservation of energy: the total energy at the beginning equals the total energy at the end. In this case, we have only potential energy at the beginning and only kinetic energy at the end. (The initial velocity is zero, and the final height is zero).

If we can find the potential energy, we can find the kinetic energy. The formula for potential energy is .

Using our given values for the mass, height, and gravity, we can solve using multiplication. Note that the height becomes negative because the book is traveling in the downward direction.

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

The kinetic energy will also equal , due to conservation of energy.

Using this value and our given mass, we can calculate the velocity from our original kinetic energy equation.

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

$22.28J=\frac{1}{2}1.12kgv^2$

$\frac{44.56J}{1.12kg}=v^2$

$39.79\frac{m^2}{s^2}=v^2$

$\sqrt{39.79\frac{m^2}{s^2}}=\sqrt{v^2}$

Since we are taking the square root, our answer can be either negative or positive. The final velocity of the book will be in the downward direction; thus, our final velocity should be negative.

$-6.31\frac{m}{s}=v$

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Question

A book falls off the top of a bookshelf. What is its final velocity right before it hits the ground?

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

Answer

The relationship between velocity and energy is:

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

We know the mass, but we need to find the total kinetic energy.

Remember the law of conservation of energy: the total energy at the beginning equals the total energy at the end. In this case, we have only potential energy at the beginning and only kinetic energy at the end. (The initial velocity is zero, and the final height is zero).

If we can find the potential energy, we can find the kinetic energy. The formula for potential energy is .

Using our given values for the mass, height, and gravity, we can solve using multiplication. Note that the height becomes negative because the book is traveling in the downward direction.

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

The kinetic energy will also equal , due to conservation of energy.

Using this value and our given mass, we can calculate the velocity from our original kinetic energy equation.

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

$95.28J=\frac{1}{2}3.23kgv^2$

$\frac{190.56J}{3.23kg}=v^2$

$58.997\frac{m^2}{s^2}=v^2$

$\sqrt{58.997\frac{m^2}{s^2}}=\sqrt{v^2}$

Since we are taking the square root, our answer can be either negative or positive. The final velocity of the book will be in the downward direction; thus, our final velocity should be negative.

$-7.68\frac{m}{s}=v$

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Question

A baseball has a mass of $\small 145g$ . If a high school pitcher can throw a baseball at $\small 38\frac{m}{s}$ , what is the approximate kinetic energy associated with this pitch?

Answer

To solve for the kinetic energy, we will need to use the equation:

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

Before we can plug in our given values, we must convert the mass from grams to kilograms. Remember, the SI unit for mass is kilograms, so most calculations will require this conversion.

$145g*\frac{1 kg}{1000g}=0.145kg$

Now we can use this mass and the given velocity to solve for the kinetic energy.

$KE=\frac{1}{2}(0.145kg)(38\frac{m}{s})^2$

$KE=\frac{1}{2}(0.145kg)(1444\frac{m^2}{s^2})$

$KE=104.69J\approx105J$

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Question

An orange falls off of a tree tall. What is the final velocity of the orange before it hits the ground?

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

Answer

For this problem, we need to use the law of conservation of energy. Since we are only looking at gravitational energy here and the orange starts at rest, we can say that the initial potential energy is equal to the final kinetic energy.

.

From here, we can expand the equation, using the formulas for gravitational potential energy and kinetic energy.

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

Notice that the mass cancels out form both sides.

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

For the height, make sure to keep that value negative as we are measuring the DISPLACEMENT rather than the distance travelled. Since displacement is a vector (magnitude and direction) we need to be clear that it travels down .

$(-9.8\frac{m}{s^2})(-5.5m)=\frac{1}{2}v^2$

$53.9\frac{m^2}{s^2}=\frac{1}{2}v^2$

$2*53.9\frac{m^2}{s^2}=v^2$

$107.8\frac{m^2}{s^2}=v^2$

$\sqrt{107.8\frac{m^2}{s^2}}=\sqrt{v^2}$

At this point, we need to remember that the square root of a positive number can be either positive or negative. Our velocity is a vector, so we will need to make sure we pick the answer choice with the appropriate direction. Since the orange is traveling downward, we know our final velocity must be negative.

$-10.38\frac{m}{s}=v$

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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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