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MCAT Physical : Potential Energy (Gravitational and Spring)

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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:

$25\frac{N}{m}$

$20\frac{N}{m}$

$80\frac{N}{m}$

$40\frac{N}{m}$

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

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

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

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

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Example Question #221 : Mcat Physical Sciences

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:

35.4cm/s

12.5cm/s

10.2cm/s

50cm/s

25cm/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:

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

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

$.125 = v^{2}$

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

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Example Question #1 : 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:

7.0J

5.3*10^-^3J

3.0J

0.1J

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.

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

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

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:

$89.3\frac{N}{m}$

$4.4\frac{N}{m}$

$3.2\frac{N}{m}$

$21.7\frac{N}{m}$

$44.4\frac{N}{m}$

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

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

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Example Question #1 : 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:

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

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

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

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

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

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

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

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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:

480m

950m

1.8km

60m

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.

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Example Question #31 : Work, Energy, And Power

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:

200J

50J

98J

75J

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.

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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.

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Example Question #1 : 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:

50kJ

20kJ

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

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Example Question #2 : 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:

26kJ

26,000kJ

2.6kJ

260J

260000J

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

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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 #221 : Mcat Physical Sciences

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 #1 : Potential Energy (Gravitational And Spring)

Example Question #31 : Work, Energy, And Power

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

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

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

All MCAT Physical Resources

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