how to find kinetic energy

All High School Physics Resources

A ball is dropped from 22m above the ground. Assuming gravity is $-9.8\frac{m}{s^2}$ , what is its final velocity?

Correct answer:

$20.77\frac{m}{s}$

Explanation:

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.

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

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

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

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

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

A 0.5kg ball rolls down a hill with an initial velocity of $2.45\frac{m}{s}$ . What is its maximum velocity?

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

Correct answer:

$5.06\frac{m}{s}$

Explanation:

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.

PE_i+KE_i=PE_f+KE_f

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

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.

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

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

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

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

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

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

A book falls off of a 2.2m high table. If the book weighs 0.75kg , what will its final velocity be right before it hits the ground?

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

Correct answer:

$6.57\frac{m}{s}$

Explanation:

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.

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

16.17J=(0.375kg)v^2

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

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

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

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

A book falls off of a 2.2m high table. If the book weighs 0.75kg , what will its kinetic energy be right before it hits the ground?

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

Correct answer:

16.17J

Explanation:

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.

PE=KE

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

mgh=PE=KE

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

16.17J=PE=KE

A rock is dropped from a helicopter hovering at 12m above the ground. If the rock weighs 0.2kg , what is its kinetic energy right before it hits the ground?

Correct answer:

23.52J

Explanation:

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

PE_i=KE_f

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

mgh=PE_i=KE_f

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.

23.52J=PE=KE

A pendulum with string length 1.2m is dropped from rest. If the mass at the end of the pendulum is 2.03kg , what is its maximum velocity?

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

Correct answer:

$4.85\frac{m}{s}$

Explanation:

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.

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

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

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

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

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

A 3.23kg book falls off the top of a 3.01m bookshelf. What is its final velocity right before it hits the ground?

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

Correct answer:

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

Explanation:

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).

PE_i=KE_f

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

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.

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

PE=95.28J

The kinetic energy will also equal 95.28J , due to conservation of energy.

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

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

$95.28J=\frac{1}{2}*3.23kg*v^2$

190.56J=3.23kg*v^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.

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

A 1.12kg book falls off the top of a 2.03m bookshelf. What is its velocity right before it hits the ground?

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

Correct answer:

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

Explanation:

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.

PE_i=KE_f

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

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.

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

PE=22.28J

The kinetic energy will also equal 22.28J , due to conservation of energy.

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

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

$22.28J=\frac{1}{2}*1.12kg*v^2$

44.56J=1.12kg*v^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.

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

An orange falls off of a tree 5.5m tall. What is the final velocity of the orange before it hits the ground?

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

Correct answer:

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

Explanation:

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.

PE_i=KE_f .

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 5.5m .

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

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

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

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

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

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.

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

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?

Correct answer:

105J

Explanation:

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.

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

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

$KE=104.69J\approx105J$

All High School Physics Resources

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