GED Science : Motion and Mechanics

Study concepts, example questions & explanations for GED Science

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Example Questions

Example Question #1 : Newton's Laws

Which of the following best illustrates Newton's Second Law?

Possible Answers:

If an object is not accelerating, then the net force on it is zero

A ball is rolled across a table and does not stop until it hits an area with friction

When two cymbals collide, some kinetic energy is converted into sound

As a hammer hits a nail, the nail also exerts a force on the hammer

Correct answer:

If an object is not accelerating, then the net force on it is zero

Explanation:

Newton's Second Law is best represented as an equation, in which the product of mass and acceleration is equal to net force:

\(\displaystyle F_{net}=ma\)

By this principle, if there is zero acceleration, then the force must also be zero.

\(\displaystyle F_{net}=m(0\frac{m}{s^2})=0N\)

A non-zero acceleration must be present if there is a non-zero net force.

Example Question #2 : Newton's Laws

Which of the following is the best example of Newton's Third Law?

Possible Answers:

A spinning top will not fall unless there is friction with the surface on which it spins

The force of being hit by an object is determined by its mass and acceleration

When two objects collide, the total initial and total final momentum will be equal

If an object's weight rests on the floor, then the floor will exert a normal force on the object

Correct answer:

If an object's weight rests on the floor, then the floor will exert a normal force on the object

Explanation:

Newton's Third Law states that for every action there is an equal and opposite reaction. In other words, a force will act in equal and opposite directions on the two objects involved.

Weight and normal force are a specific example of this principle. Acceleration due to gravity creates a downward force, known as weight. This force, however, results in zero downward acceleration when the object is resting on a surface (a book does not fall through a table, for example). For the forces to be in equilibrium, there must be a counteracting upward force. This is known as the normal force, which is equal and opposite to the weight of the object. While weight pulls the object down, normal force pushes it up to create equilibrium.

Example Question #1 : Physics

A car drives \(\displaystyle \small 5\) miles east and \(\displaystyle \small 12\) miles north. What is the total displacement of the car?

Possible Answers:

\(\displaystyle 17\ \text{miles}\)

\(\displaystyle 13\ \text{miles}\)

\(\displaystyle 7\ \text{miles}\)

\(\displaystyle 60\ \text{miles}\)

Correct answer:

\(\displaystyle 13\ \text{miles}\)

Explanation:

The car has covered a total distance of \(\displaystyle \small 17\) miles, however, displacement and distance are not the same thing. Distance is simply the measure of how far the car have moved, regardless of direction. Displacement is a vector quantity, and takes direction into account. Displacement measures the total distance only between the car's starting point and ending point.

To solve this question, it is best to draw a sketch. The resulting sketch will be a right triangle with the hypotenuse equal to the displacement. The Pythagorean Theorem can be used to solve the problem.

Question_1

\(\displaystyle \small a^2+b^2=c^2\)

\(\displaystyle \small 5^2+12^2=d^2\)

\(\displaystyle \small 25+144=d^2\)

\(\displaystyle \small 169=d^2\)

\(\displaystyle \small \sqrt{169}=\sqrt{d^2}\)

\(\displaystyle \small 13=d\)

Example Question #1 : Motion And Mechanics

A car travels due east for two hours and covers 144km. What is the car's average velocity in meters per second?

Possible Answers:

\(\displaystyle \small 72000\frac{m}{s}\)

\(\displaystyle \small 72\frac{m}{s}\)

\(\displaystyle \small 0.05\frac{m}{s}\)

\(\displaystyle \small \small 20\frac{m}{s}\)

Correct answer:

\(\displaystyle \small \small 20\frac{m}{s}\)

Explanation:

This question requires us to convert between units, then solve for the velocity.

First, convert kilometers to meters. There are 1000m in 1km.

\(\displaystyle 144km\times \frac{1000m}{1km}=144000m\)

Then, convert hours to seconds. There are 60min in 1hr, and 60sec in 1min. In other words, there are 3600s in 1hr.

\(\displaystyle 2hr\times \frac{60min}{1hr}\times \frac{60s}{1hr}=7200s\)

Now that you have both the distance in meters and the time in seconds, you can find the velocity. Velocity is equal to the distance divided by the time.

\(\displaystyle v=\frac{d}{t}\)

\(\displaystyle v=\frac{144000m}{7200s}=20\frac{m}{s}\)

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