MCAT Physical : Forces

Study concepts, example questions & explanations for MCAT Physical

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Example Questions

Example Question #21 : Forces

Using Newton's Law of Universal Gravitation equation, which of the following expressions is equal to the local gravitational acceleration  on Earth?

Possible Answers:

Correct answer:

Explanation:

On earth, .

The law of universal gravitation is equal to .

We can set these equations equal to one another and isolate  by dividing both sides by , the mass of an object on Earth.

Using the mass of the Earth, the radius of the Earth, and the gravitational constant, , we get a value of approximately if we solve for .

Example Question #21 : Forces

Which of the following must be true for a  block to be considered weightless?

Possible Answers:

The block is being constantly accelerated downward at 

The block is being constantly accelerated downward at 

The block is being constantly accelerated downward at  

The object can never be considered weightless

The block is being constantly accelerated upward at 

Correct answer:

The block is being constantly accelerated downward at 

Explanation:

One way an object can be considered "weightless," is if it is accelerating downward at the same rate that it would be if it were free falling. This is why objects in an elevator whose cable had been cut would be "weightless." Therefore the answer is just for the block to be accelerating in the same magnitude and direction as the acceleration due to gravity, which is 

Example Question #2 : Universal Gravitation

Which of the following changes would increase a satellite's orbital speed?

Possible Answers:

Increase the mass of the satellite

Decrease the mass of the satellite

Decrease orbital radius

Increase orbital radius

Decrease the mass of the Earth

Correct answer:

Decrease orbital radius

Explanation:

We know , and the force of gravity is:

 

Also, the equation for uniform circular motion, such as a satellite in orbit is:

 

Set  and substitute the acceleration due to circular motion into the equation. Solve for velocity.

This indicates that the only variables that affect the orbital speed are orbital radius and the mass of the Earth.

Example Question #1 : Gravity And Weight

block lies upon a frictionless surface. A string is attached to the right side of the block, passed over a pulley, and then attached to a  mass suspended by the string. When the block is held still, what is the tension in the string?

Possible Answers:

Correct answer:

Explanation:

The block cannot provide any frictional force to the left because it is resting on a frictionless table. Whether the block is allowed to move or not, the tension force in the string is all generated by the suspended mass. The force in the string is equal and opposite to the gravitational force acting on the suspended mass.

The system is in equilibrium when no parts are in motion. Alternatively stated, the upward tension in the string is equal and opposite to the downwards force on the mass. If the block is not allowed to move as more mass is added, at some point the tensile force in the string will exceed its mechanical limits, and the string will break.

Example Question #1 : Gravity And Weight

What is the tension in a rope that is used to pull a  box straight upwards with an acceleration of

Possible Answers:

Correct answer:

Explanation:

For the problem we need to understand Newton's second law: . The net upward forces and net downward forces must equal the product of mass and acceleration. Since the box is traveling upwards with an acceleration of 1m/s2 we can indicate upward forces as positive and downward forces as negative. Indicating T as tension and Fg as the weight of the box, we can find the value of tension with the equation .

Example Question #3 : Gravity And Weight

Planet X has mass  and radius , and the gravitational acceleration on its surface is . What is the gravitational acceleration at the location of a satellite orbiting at a distance  from the surface of planet X? 

Possible Answers:

Correct answer:

Explanation:

Gravitational acceleration is related to distance via the equation:

 is the gravitational constant,  is the mass of the attracting object, and  is the distance from its center.

In this case, the initial distance (on the surface) is from the center of planet X, and the distance of the satellite is  from the center. We know that gravitational acceleration is proportional to the distance squared, and we know the acceleration at the surface. Using these values, we can solve for the acceleration on the satellite.

By expanding the equation for the acceleration on the satellite, we can see that it is equal to one-sixteenth the acceleration at the surface, based on our original equation. Substitute the value of the surface acceleration to get the final answer.

 

Example Question #1 : Gravity And Weight

A ball is thrown vertically with an initial velocity, , and returns to its original position after time . How would the value of be affected if the ball were thrown in the same manner on the moon, where gravitational acceleration is one-sixth the gravitational acceleration on Earth?

Possible Answers:

Decrease by a factor of six

Increase by a factor of six

Increase by a factor of 

Decrease by a factor of 

Correct answer:

Increase by a factor of six

Explanation:

To solve this question, we will need to use the equation for acceleration:

In this case, the initial velocity will be equal to the final velocity, but opposite in magnitude. The initial velocity is in the upward direction, while the final velocity is downward.

Plug this value into the equation for acceleration.

The velocity value is constant, regardless of the planet. Substitute the acceleration for each planet to determine the change in the time variable.

We can see that the time of flight on the moon is equal to six times the time of flight on Earth.

Example Question #2 : Gravity And Weight

A hypothetical planet has a radius equal to twice that of Earth, with the same mass as Earth. How much would a person weigh on the surface of this hypothetical planet if they weighed 1000N on earth?

Possible Answers:

Correct answer:

Explanation:

The force due to gravity on any object can be given by the equation below.

 is the gravitational constant,  is the mass of the earth,  is the mass of the object, and  is the distance between the center of each object.

In our question, the only value to change is the radius of the new planet; both masses and  remain constant. The effect of doubling the radius on the force is given below.

The person's weight on the new planet would be one-fourth their weight on Earth.

Example Question #3 : Gravity And Weight

An astronaut standing on a scale on the moon observes that he weighs . If the acceleration due to gravity on the moon's is one-sixth of its value on Earth's surface, what is the astronaut's mass on the earth?

Possible Answers:

Correct answer:

Explanation:

To relate force and mass, we use Newton's second law:

We are given his weight (force) on the moon, and we are told the relative gravitational acceleration on the moon. Using these values, we can find the astronaut's mass.

Since mass is the same regardless of gravitational acceleration, this is the same as the astronaut's mass on the earth.

Example Question #30 : Forces

A person stands on a scale in an elevator. When the elevator is moving upwards at a constant velocity of 0.5m/s, the scale reads 500N. If the elevator then slows down, with a deceleration of 0.5m/s2, what is the new reading?

Possible Answers:

475N

575N

375N

425N

525N

Correct answer:

475N

Explanation:

It may be helpful to start with a free-body diagram showing the forces acting on the person. We have the gravitational force, Fg, downwards, and the normal force of the scale, Fn, upwards.

Use these to write the net force equation.

First we need to solve for the person’s mass, m. When the elevator is moving at a constant rate, we are given that Fn = 500N, and we can solve for m.

.

When the elevator comes to rest, then we have an acceleration of -0.5 m/s2 (downwards acceleration). Plugging this in, we can solve for the new Fn.

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