Call Now to Set Up Tutoring:

(888) 888-0446

MCAT Physical : Potential Energy (Gravitational and Spring)

Study concepts, example questions & explanations for MCAT Physical

Create An Account

All MCAT Physical Resources

8 Diagnostic Tests 303 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

← Previous 1 2 Next →

Example Question #1 : Potential Energy (Gravitational And Spring)

A spring in a pinball machine is compressed 0.5m from its equilibrium position. It is released and it fires the 0.1kg pinball at a velocity of $10\frac{m}{s}$ into play. No energy is lost to friction and all of the energy in the spring is transferred to the pinball. What is the spring constant in the spring?

Possible Answers:

$40\frac{N}{m}$

$20\frac{N}{m}$

$25\frac{N}{m}$

$80\frac{N}{m}$

Correct answer:

$40\frac{N}{m}$

Explanation:

The energy in a spring is given by the equation $U=\frac{1}{2}kx^{2}$ , where is the spring constant and is the displacement from the equilibrium position. The energy is transferred into kinetic energy in the pinball.

The kinetic energy in the pinball is found using the equation $KE= \frac{1}{2}mv^{2}$ .

$KE= \frac{1}{2}(0.1 kg)(10\frac{m}{s})^{2}$

$KE= (0.5)(0.1kg)(100\frac{m^2}{s^2})$

KE= 5 J

Now that we have kinetic energy, we can find the equal energy that was stored in the spring to begin.

U=KE

$\frac{1}{2}kx^{2}=\frac{1}{2}mv^2=5J$

$5 J=\frac{1}{2}(k)(0.5m)^{2}$

$k= \frac{(2)(5J)}{(0.5m)^{2}} = \frac{10J}{0.25m^2}=40\frac{N}{m}$

Report an Error

Example Question #2 : Potential Energy (Gravitational And Spring)

A 2kg mass is attached to a massless spring with force constant 100N/m. The system rests on a frictionless horizontal surface. If the spring is compressed 5cm, then released, what is the maximum velocity of the mass?

Possible Answers:

25cm/s

50cm/s

35.4cm/s

12.5cm/s

10.2cm/s

Correct answer:

35.4cm/s

Explanation:

Use conservation of energy to compare the point of maximum compression and the point of maximum velocity. At maximum compression, all energy is spring potential energy, where $E_{s}= \frac{1}{2}kx^{2}$ , and at maximum velocity all energy is kinetic, where $E_{k}= \frac{1}{2}mv^{2}$ . Since energy must be conserved, these two energies must be equal: $\frac{1}{2}kx^{2} = \frac{1}{2}mv^{2}$ .

Plugging in the given information for E_s and solving for v gives the following series of calculations:

$E_s=\frac{1}{2}kx^{2} = \frac{1}{2}(100)(0.05)^{2}=.125 Joules$

$.125 Joules = \frac{1}{2}mv^{2}$

$.125 = v^{2}$

$v = .354 m/s = 35.4 cm/s$

Report an Error

Example Question #3 : Potential Energy (Gravitational And Spring)

If a ball is attached to a spring that has a spring constant of 20N/m and is compressed 0.1m, what is the total energy of the oscillation?

Possible Answers:

0.1J

5.3*10^-^3J

3.0J

7.0J

17.0J

Correct answer:

0.1J

Explanation:

The total energy of a simple harmonic oscillating system can be determined by the equation $E=\frac{1}{2}kA^2$ . A is the amplitude of the oscillation, k is the spring constant, and E is the total energy. Plugging in values given by the equation, the total energy is equal to 0.1J.

$E=\frac{1}{2}(20\frac{N}{m})(0.1m)=1J$

Report an Error

Example Question #221 : Mcat Physical Sciences

A massless spring rests horizontally on a table, with its right end attached to a $\small 250g$ block and with its left end attached to a wall. The block is then pushed against the spring, compressing it $\small 15cm$ from its equilibrium position. After being released, the block's velocity as it moves back through the equilibrium point is $\small 2\frac{m}{s}$ . What is the spring constant of this spring?

Possible Answers:

$89.3\frac{N}{m}$

$4.4\frac{N}{m}$

$21.7\frac{N}{m}$

$44.4\frac{N}{m}$

$3.2\frac{N}{m}$

Correct answer:

$44.4\frac{N}{m}$

Explanation:

Use conservation of energy to equate the initial spring potential energy and the final kinetic energy:

$K_{i} +U_{i}=K_{f}+U_{f}$

The initial situation is defined when the spring and block are compressed, and the final situation is defined to be when the block moves back through the equilibrium point. For a spring, the potential energy is maximal when the spring is compressed and the kinetic energy is maximal when the mass passes equilibrium. We can set up a equation based on these factors.

$K_i=0\ \text{and}\ U_f=0$

U_i=K_f

Use the equations for spring potential energy and for kinetic energy to incorporate the appropriate variables into the equation.

$U=\frac{1}{2}kx^2\ \text{and}\ KE=\frac{1}{2}mv^2$

$\frac{1}{2}kx_i^2=\frac{1}{2}mv_f^2$

We are given values for the displacement, mass, and final velocity. Using these values, we can isolate the spring constant.

$\frac{1}{2}k(0.15m)^2=\frac{1}{2}(0.25kg)(2\frac{m}{s})^2$

(0.01125m^2)k=0.5J

$k=44.4\frac{N}{m}$

Report an Error

Example Question #5 : Potential Energy (Gravitational And Spring)

A block with mass, , is attached to a spring with a spring constant, . The spring is compressed a maximum distance, , and released. What is the maximum velocity the block reaches after the spring is released?

Possible Answers:

$\nu = \frac{1}{2}kx^{2}m$

$\nu =\frac{ \frac{1}{2}kx^{2}}{m}$

$\nu = \sqrt{\frac{kx^{2}}{m}}$

$\nu = \sqrt{kx^{2}m}$

Correct answer:

$\nu = \sqrt{\frac{kx^{2}}{m}}$

Explanation:

The spring potential is given by the equation $U = \frac{1}{2}kx^{2}$ and the equation for kinetic energy is given by $KE = \frac{1}{2}mv^{2}$ .

When is at a maximum distance (the amplitude) the velocity is zero, and the spring ONLY has potential energy. When it is released, the total energy is part kinetic and part potential. When the amplitude is at zero (equilibrium), the velocity is at a maximum and all the potential energy is converted into kinetic energy. We can then set the two equations equal to each other to solve the maximum velocity.

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

kx^2=mv^2

$\frac{kx^2}{m}=v^2$

$\nu = \sqrt{\frac{kx^{2}}{m}}$

Report an Error

Example Question #1 : Potential Energy (Gravitational And Spring)

A certain chocolate candy bar contains $\small 150kcal$ . How high can a $\small 20kg$ barbell be lifted with the energy contained in this candy bar?

1kcal=4184J

Possible Answers:

60m

480m

950m

1.8km

3.2km

Correct answer:

3.2km

Explanation:

Our first step is to convert kilocalories to Joules.

$150kcal*\frac{4184J}{1kcal}=627,600J=627.6kJ$

Now that we have converted to standard units, we can use the formula for gravitational potential energy to find the height.

PE=mgh

We know our energy limit from the candy bar, the mass of the barbell, and the gravitational acceleration.

$627.6kJ=(20kg)(9.8\frac{m}{s^2})h$

Solve to isolate the height.

$h=\frac{627.6kJ}{(20kg)(9.8\frac{m}{s^2})}$

h=3.2km

Note that we end up with kilometers because we used kilojoules in our calculation.

Report an Error

Example Question #1 : Potential Energy (Gravitational And Spring)

A 2kg mass is suspended on a rope that wraps around a frictionless pulley attached to the ceiling with a mass of 0.01kg and a radius of 0.25m. The other end of the rope is attached to a massless suspended platform, upon which 0.5kg weights may be placed. While the system is initially at equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling on the rope.

Screen_shot_2013-10-09_at_10.32.21_pm

After the rope is cut and the platform falls, we want to pull the platform back up to 5m above the ground. How much energy is required to raise the platform?

Possible Answers:

75J

200J

98J

50J

Correct answer:

98J

Explanation:

First, we need to understand what type of energy we are considering. As the resistive force to motion is due to gravity, we are talking about gravitational potential energy. We need to use the formula U = mgh.

Plugging in the values that are provided, we can solve for the potential energy (U).

U = (2kg)(9.8m/s²)(5m) = 98J

The amount of work required is equal to the change in potential energy of the platform.

Report an Error

Example Question #1 : Potential Energy (Gravitational And Spring)

Two students (student X and student Y) lift a boulder vertically from point A to point B. Student X directly lifts the boulder from point A to point B, whereas student Y uses a pulley to lift the boulder. This allows student Y to apply a force ( F_y ) that is three times smaller than the force applied by student X ( F_x ). Both students apply force upwards and take the same amount of time to complete this task.

The vertical distance between point A and point B is .

Student Z uses an inclined plane to lift the boulder and exerts only one third of F_x . Which of the following quantities is not equal for student Z and student X?

I. The change in potential energy

II. The distance travelled

III. The sum of the change in kinetic energy and change in potential energy

Possible Answers:

I only

II only

I and II

I and III

Correct answer:

II only

Explanation:

The change in potential energy is defined as:

$\Delta PE = mg\Delta h$

This means that the potential energy is only dependent on the change in vertical height, which is the same for both students. Since both students travel the same distance, the change in potential energy is equal and statement I is true.

Student X directly lift the boulder whereas student Z uses an inclined plane. Student Z slides the boulder along a larger distance than student X, even if their total displacement ends up begin equal; therefore, statement II is false.

Recall that the sum of change in kinetic energy and change in potential energy is equal to work, which is the same for both students (work is path independent in a gravitational field). The total work for both students will be equal, and statement III will be true.

The only statement that is not equal for both students is statement II.

Report an Error

Example Question #9 : Potential Energy (Gravitational And Spring)

A 2kg lead ball is loaded into a spring cannon and the cannon is set at a 45º angle on a platform. The spring has a spring constant of 100N/m and the ball and spring system is compressed by 1m before launch. While the ball is in flight air resistance can be neglected, and the ball finishes its flight by landing at a cushion placed some distance away from the cannon.

How much energy is stored in the spring before the ball is launched?

Possible Answers:

20kJ

50kJ

50J

20J

Correct answer:

50J

Explanation:

In order to determine how much energy is stored, we first need to understand what type of energy we want to consider. A spring stores potential energy; the potential energy of the spring is maximized at maximal displacement from its resting state. In order to compute the potential energy stored, we need both the spring constant (100N/m) and the displacement from resting (1m).

PE_s = ½k(Δx)²= ½(100N/m)(1m)²= 50J

Report an Error

Example Question #1 : Potential Energy (Gravitational And Spring)

Two children are playing with sleds on a snow-covered hill. Sam weighs 50kg, and his sled weighs 10kg. Sally weighs 40kg, and her sled weighs 12kg. When they arrive, they climb up the hill using boots. Halfway up the 50-meter hill, Sally slips and rolls back down to the bottom. Sam continues climbing, and eventually Sally joins him at the top.

They then decide to sled down the hill, but disagree about who will go first.

Scenario 1:

Sam goes down the hill first, claiming that he will reach a higher velocity. If Sally had gone first, Sam says they could collide.

Scenario 2:

Sally goes down the hill first, claiming that she will experience lower friction and thus reach a higher velocity. If Sam had gone first, Sally says they could collide.

Scenario 3:

Unable to agree, Sam and Sally tether themselves with a rope and go down together.

How much potential energy does Sally have at the top of the hill on her sled?

Possible Answers:

2.6kJ

260J

26,000kJ

260000J

26kJ

Correct answer:

26kJ

Explanation:

This is a tricky question only because you have to keep your units straight. The formula is simply PE = mgh.

PE = 52kg * 10m/s² * 50m = 26,000J = 26kJ

Report an Error

← Previous 1 2 Next →

View MCAT Chemical and Physical Foundations of Biological Systems Tutors

Yongmao
Certified Tutor

Nankai University, Bachelor of Science, Chemistry. Washington University in St Louis, Doctor of Philosophy, Organic Chemistry.

View MCAT Chemical and Physical Foundations of Biological Systems Tutors

Christopher
Certified Tutor

Rose-Hulman Institute of Technology, Bachelor of Science, Physics. University of Washington, Master of Science, Physics.

View MCAT Chemical and Physical Foundations of Biological Systems Tutors

Matt
Certified Tutor

All MCAT Physical Resources

8 Diagnostic Tests 303 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Computer Science Tutors in Dallas Fort Worth, SAT Tutors in Phoenix, Calculus Tutors in San Francisco-Bay Area, Computer Science Tutors in Phoenix, ACT Tutors in Washington DC, MCAT Tutors in San Diego, Statistics Tutors in Los Angeles, French Tutors in Philadelphia, Physics Tutors in Chicago, LSAT Tutors in New York City

Popular Courses & Classes

ISEE Courses & Classes in Phoenix, SAT Courses & Classes in Washington DC, Spanish Courses & Classes in Houston, GMAT Courses & Classes in Houston, GRE Courses & Classes in Philadelphia, GMAT Courses & Classes in San Francisco-Bay Area, GMAT Courses & Classes in Washington DC, ISEE Courses & Classes in Houston, ACT Courses & Classes in New York City, SAT Courses & Classes in Atlanta

Popular Test Prep

GMAT Test Prep in Miami, GMAT Test Prep in Philadelphia, GMAT Test Prep in Washington DC, ACT Test Prep in Boston, MCAT Test Prep in Phoenix, MCAT Test Prep in Washington DC, GMAT Test Prep in Seattle, LSAT Test Prep in Philadelphia, SSAT Test Prep in Boston, SAT Test Prep in New York City

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

MCAT Physical : Potential Energy (Gravitational and Spring)

Study concepts, example questions & explanations for MCAT Physical

All MCAT Physical Resources

Example Questions

Example Question #1 : Potential Energy (Gravitational And Spring)

Example Question #2 : Potential Energy (Gravitational And Spring)

Example Question #3 : Potential Energy (Gravitational And Spring)

Example Question #221 : Mcat Physical Sciences

Example Question #5 : Potential Energy (Gravitational And Spring)

Example Question #1 : Potential Energy (Gravitational And Spring)

Example Question #1 : Potential Energy (Gravitational And Spring)

Example Question #1 : Potential Energy (Gravitational And Spring)

Example Question #9 : Potential Energy (Gravitational And Spring)

Example Question #1 : Potential Energy (Gravitational And Spring)

All MCAT Physical Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: