MCAT Physical - 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?

50J

50kJ

20J

20kJ

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).

PEs = ½k(Δx)2 = ½(100N/m)(1m)2 = 50J

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?

$44.4\frac{N}{m}$

$3.2\frac{N}{m}$

$4.4\frac{N}{m}$

$21.7\frac{N}{m}$

$89.3\frac{N}{m}$

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.

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

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.

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

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

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?

26kJ

26,000kJ

260J

2.6kJ

260000J

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

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?

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.

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

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?

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.

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.

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

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

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?

35.4cm/s

12.5cm/s

25cm/s

10.2cm/s

50cm/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 Es and solving for v gives the following series of calculations:

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

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

$.125 = v^{2}$

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

A $\small 1000kg$ pyramid stone is raised to a height of $\small 110m$ . What is the potential energy of the system?

It depends if the stone is raised vertically or along an incline

Explanation

No matter how you raise the stone (all at once, in multiple vertical steps, or along a ramp) the final potential energy is the same. Gravitational potential energy of a given object is solely dependent on height.

Use the given mass, the acceleration of gravity, and the final height to calculate the potential energy of the system.

$PE=(1000kg)(9.8\frac{m}{s^2})(110m)$

$PE=1078000\frac{m^2kg}{s^2}=1.078*10^6J$

Since work is equal to the change in potential energy of a system, this value also gives the total work done to raise the stone.

A 2kg box slides down a ramp a distance of two meters before it reaches the ground. The ramp has an angle of 30o. The coefficient of kinetic friction for the ramp is 0.1.

What is the gravitational potential energy of the box relative to the ground when at the top of the ramp?

Explanation

Gravitational potential energy is determined using the equation $\Delta U_{g} = mgh$ , with being the mass of the object, being the gravitational acceleration, and being the height of the object relative to the ground. Because we know the length of the ramp and the angle of the ramp, we can solve for the box's height above the ground.

$sin(30^{\circ}) = \frac{x}{2}$

$0.5=\frac{x}{2}$

Now that we have the height, we simply plug the given values into the equation.

$\Delta U_{g} = (2kg)(10\frac{m}{s^{2}})(1m) = 20J.$

A spring in a pinball machine is compressed from its equilibrium position. It is released and it fires the 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?

$40\frac{N}{m}$

$20\frac{N}{m}$

$80\frac{N}{m}$

$25\frac{N}{m}$

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

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

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

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

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

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

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

A spring is used to launch a $\small 1kg$ ball straight up to a maximum height of $\small 5m$ . The spring constant of the spring is $4\frac{N}{m}$ .

What was the displacement of the spring required to launch the ball?

Explanation

We can use conservation of energy to calculate the displacement. The spring must be displaced so that its potential energy is equal to the gravitational potential energy of the ball at its maximum height.

$PE_{spring} = PE_{grav}$

Substituting in the expressions for potential energy of the spring and gravitational potential energy, we can start to set up an equation using the given variables.

$PE_{spring}=\frac{1}{2}kx^2\ \text{and}\ PE_{grav}=mgh$

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

We are given the spring constant, the mass, and the maximum height. Using these values, and the acceleration for gravity, we can solve for the spring displacement.

$\frac{1}{2}(4\frac{N}{m})x^2=(1kg)(9.8\frac{m}{s^2})(5m)$

$(2\frac{N}{m})x^2=49J$

$x=4.95m\rightarrow -4.95m$

Note that, since we are taking the square root, the displacement answer could be either positive or negative. Since the spring is being compressed to launch the ball (rather than stretched), the displacement should be negative.

Potential Energy (Gravitational and Spring)

Help Questions

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation

Explanation