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SAT II Physics : Work, Energy, and Power

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Example Questions

Example Question #1 : Work, Energy, And Power

A 4kg cart has a linear momentum with a magnitude of $20\frac{kg\cdot m}{s}$ . What is the cart's kinetic energy?

Possible Answers:

30J

10J

100J

20J

50J

Correct answer:

50J

Explanation:

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

p=mv

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.

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

KE=50J

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Example Question #31 : Energy

A 2kg book falls off the top of a 2.3m bookshelf. What is its velocity right before it hits the ground?

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

Possible Answers:

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

$13.43\frac{m}{s}$

$3.36\frac{m}{s}$

$-26.86\frac{m}{s}$

$45.08\frac{m}{s}$

Correct answer:

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

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.

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

PE=45.08J

The kinetic energy will also equal 45.08J , due to conservation of energy.

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

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

$45.08J=\frac{1}{2}*2kg*v^2$

45.08J=1kg*v^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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Example Question #8 : Calculating Work

A 30kg crate is pushed 15m across the floor. If 19N of force was used to achieve this motion, how much work was done?

Possible Answers:

5586J

285J

1.3J

30J

Correct answer:

285J

Explanation:

The formula for work is:

W=F*d

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

W=19N*15m

W=285J

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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Example Question #1 : Work

A man moves a 40kg crate by pulling a rope at a 60^o angle with the ground and a constant force of 300N . How much work has the man performed after pulling the crate 10m ?

Possible Answers:

1500J

3000J

2500J

2000J

1800J

Correct answer:

1500J

Explanation:

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.

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

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

W=1500J

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SAT II Physics : Work, Energy, and Power

Study concepts, example questions & explanations for SAT II Physics

All SAT II Physics Resources

Example Questions

Example Question #1 : Work, Energy, And Power

Example Question #31 : Energy

Example Question #8 : Calculating Work

Example Question #1 : Work

All SAT II Physics Resources

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