Work, Energy, and Power - SAT

Card 0 of 16

Question

A crate is pushed across the floor. If of force was used to achieve this motion, how much work was done?

Answer

The formula for work is:

Given the values for force and distance, we can calculate the work done.

Note that no work is done by the force of gravity or the weight of the box, since the vertical position does not change.

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Question

A man moves a crate by pulling a rope at a angle with the ground and a constant force of . How much work has the man performed after pulling the crate ?

Answer

Work is given by the equation:

$W=Fd\cos\theta$

We know the force with which the man pulls, the angle, and the distance that the crate travels. Using these values we can solve the equation for the work done.

$W=(300N)(10m)\cos(60^o)$

$W=(3000J)(\frac{1}{2})$

Compare your answer with the correct one above

Question

A crate is pushed across the floor. If of force was used to achieve this motion, how much work was done?

Answer

The formula for work is:

Given the values for force and distance, we can calculate the work done.

Note that no work is done by the force of gravity or the weight of the box, since the vertical position does not change.

Compare your answer with the correct one above

Question

A man moves a crate by pulling a rope at a angle with the ground and a constant force of . How much work has the man performed after pulling the crate ?

Answer

Work is given by the equation:

$W=Fd\cos\theta$

We know the force with which the man pulls, the angle, and the distance that the crate travels. Using these values we can solve the equation for the work done.

$W=(300N)(10m)\cos(60^o)$

$W=(3000J)(\frac{1}{2})$

Compare your answer with the correct one above

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

A cart has a linear momentum with a magnitude of $20\frac{kg\cdot m}{s}$ . What is the cart's kinetic energy?

Answer

Linear momentum is calculated as the product of mass and velocity:

We are given the mass of the cart and its momentum, allowing us to solve for its velocity.

$20\frac{kg\cdot m}{s}=(4kg)v$

$v=\frac{20\frac{kg\cdot m}{s}}{4kg}=5\frac{m}{s}$

Now that we know the velocity of the cart, we need to use the equation for kinetic energy:

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

Use the value of the velocity and the given mass of the cart to solve.

$KE=\frac{1}{2}(4kg)(5\frac{m}{s})^2$

Compare your answer with the correct one above

Question

A crate is pushed across the floor. If of force was used to achieve this motion, how much work was done?

Answer

The formula for work is:

Given the values for force and distance, we can calculate the work done.

Note that no work is done by the force of gravity or the weight of the box, since the vertical position does not change.

Compare your answer with the correct one above

Question

A man moves a crate by pulling a rope at a angle with the ground and a constant force of . How much work has the man performed after pulling the crate ?

Answer

Work is given by the equation:

$W=Fd\cos\theta$

We know the force with which the man pulls, the angle, and the distance that the crate travels. Using these values we can solve the equation for the work done.

$W=(300N)(10m)\cos(60^o)$

$W=(3000J)(\frac{1}{2})$

Compare your answer with the correct one above

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.

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

Compare your answer with the correct one above

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.

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

Compare your answer with the correct one above

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.

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

Compare your answer with the correct one above

Question

A cart has a linear momentum with a magnitude of $20\frac{kg\cdot m}{s}$ . What is the cart's kinetic energy?

Answer

Linear momentum is calculated as the product of mass and velocity:

We are given the mass of the cart and its momentum, allowing us to solve for its velocity.

$20\frac{kg\cdot m}{s}=(4kg)v$

$v=\frac{20\frac{kg\cdot m}{s}}{4kg}=5\frac{m}{s}$

Now that we know the velocity of the cart, we need to use the equation for kinetic energy:

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

Use the value of the velocity and the given mass of the cart to solve.

$KE=\frac{1}{2}(4kg)(5\frac{m}{s})^2$

Compare your answer with the correct one above

Question

A cart has a linear momentum with a magnitude of $20\frac{kg\cdot m}{s}$ . What is the cart's kinetic energy?

Answer

Linear momentum is calculated as the product of mass and velocity:

We are given the mass of the cart and its momentum, allowing us to solve for its velocity.

$20\frac{kg\cdot m}{s}=(4kg)v$

$v=\frac{20\frac{kg\cdot m}{s}}{4kg}=5\frac{m}{s}$

Now that we know the velocity of the cart, we need to use the equation for kinetic energy:

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

Use the value of the velocity and the given mass of the cart to solve.

$KE=\frac{1}{2}(4kg)(5\frac{m}{s})^2$

Compare your answer with the correct one above

Question

A cart has a linear momentum with a magnitude of $20\frac{kg\cdot m}{s}$ . What is the cart's kinetic energy?

Answer

Linear momentum is calculated as the product of mass and velocity:

We are given the mass of the cart and its momentum, allowing us to solve for its velocity.

$20\frac{kg\cdot m}{s}=(4kg)v$

$v=\frac{20\frac{kg\cdot m}{s}}{4kg}=5\frac{m}{s}$

Now that we know the velocity of the cart, we need to use the equation for kinetic energy:

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

Use the value of the velocity and the given mass of the cart to solve.

$KE=\frac{1}{2}(4kg)(5\frac{m}{s})^2$

Compare your answer with the correct one above

Question

A crate is pushed across the floor. If of force was used to achieve this motion, how much work was done?

Answer

The formula for work is:

Given the values for force and distance, we can calculate the work done.

Note that no work is done by the force of gravity or the weight of the box, since the vertical position does not change.

Compare your answer with the correct one above

Question

A man moves a crate by pulling a rope at a angle with the ground and a constant force of . How much work has the man performed after pulling the crate ?

Answer

Work is given by the equation:

$W=Fd\cos\theta$

We know the force with which the man pulls, the angle, and the distance that the crate travels. Using these values we can solve the equation for the work done.

$W=(300N)(10m)\cos(60^o)$

$W=(3000J)(\frac{1}{2})$

Compare your answer with the correct one above

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