### All AP Physics 1 Resources

## Example Questions

### Example Question #901 : Newtonian Mechanics

A block with a mass of 1kg is hanging vertically from a spring that has a constant of . If the block is pulled downward 1.5m past equilibrium and released, what is the velocity of the block as it passes back through the equilibrium position?

**Possible Answers:**

**Correct answer:**

This problem deals with the conservation of energy in the form of a spring. The formula for the conservation of energy is:

We can eliminate initial kinetic energy to get:

We have two forms of potential energy in this problem: gravitational and spring. Therefore, we can rewrite the expression to be:

There is no initial gravitational potential energy and no final spring potential energy. Subsituting our expressions for potential and kinetic energy, we get:

Rearranging for final velocity, we get:

is the original displacement (the same as ).

We have all of the values, so we can solve:

### Example Question #2 : Springs

A spring with a constant of is hanging from the ceiling of an elevator. A block of mass 1kg is attached to the end of the spring. If the elevator begins to accelerate at a rate of , how far will the equilibrium position of the spring shift?

**Possible Answers:**

**Correct answer:**

Before the elevator begins to accelerate, the block has a weight of:

While the elevator is accelerating, the block has a weight of:

Now we can say that the weight of the block has effectively increased by 2N.

We can use this increase in force to find out how much the equilibrium position will shift:

Rearranging for x (displacement), we get:

### Example Question #3 : Springs

A horizontal spring with a constant of is on a frictionless surface. If a mass of attached to the spring has a maximum acceleration of , what is its maximum velocity?

**Possible Answers:**

**Correct answer:**

We know that the maximum acceleration of a spring occurs when it is at its maximum displacement. At this point, we can write an expression for the force that the block is under:

Rearranging for the distance, we get:

This is also the point at which the spring is storing its maximum potential energy:

We can substitute our rearranged distance equation into the equation for potential energy:

We also know that the maximum velocity of a spring occurs when it is at the point of equilibrium. Since the spring is on a frictionless surface, we can say that its kinetic energy at this point is equal to the maximum potential energy:

Plugging in the expressions for these variables:

Rearranging for velocity:

We know all of the values, so we can plug and chug:

### Example Question #1 : Circular, Rotational, And Harmonic Motion

A horizontal spring with a spring constant of sits on a table and has a mass of attached to one end. The coefficient of kinetic friction between the mass and table is . If the spring is stretched to a distance of past its point of equilibrium and released, how many times does the mass pass through the point of equilibrium before coming to rest?

**Possible Answers:**

**Correct answer:**

From the problem statement, we can calculate how much potential energy is initially stored in the spring. Since we are neglecting air resistance, we can say that all of this energy is lost through friction as the system comes to a stop.

Calculating initial potential energy:

Then we can write an expression for the force of friction on the mass:

We can then write an expression of the work done by friction:

Setting this equal to the initial potential energy:

Therefore, the mass travels a total distance of 20 meters before coming to rest. We know that the distance between maximum compression and maximum extension of the spring is 4 meters. Therefore, we can say that the mass travels from one maximum displacement to the other five times, passing through the equilibrium once each time. The mass passes through the equilibrium point 5 times.

### Example Question #951 : Ap Physics 1

A spring with a spring constant of is compressed past its point of equilibrium. If internal friction results in an average power loss of , how long does the spring oscillate after being released until it comes to rest?

**Possible Answers:**

**Correct answer:**

We can calculate the initial potential energy of the spring from what we are given in the problem statement:

Assuming the only energy loss of the spring is through internal friction, we can write:

### Example Question #6 : Springs

A block falls off of a table and collides with a spring on the floor with velocity . The spring has a spring constant of . How much is the spring compressed when the velocity of the block is ?

**Possible Answers:**

**Correct answer:**

This problem can be solved by the conservation of energy. The block initially has kinetic energy equal to:

When the velocity of the block is zero, all the kinetic energy has been transferred to potential spring energy.

### Example Question #7 : Springs

A homogenous mass of 0.25kg is fixed to a 0.5kg Hookean spring. When the mass/spring system is stretched 1cm from the equilibrium, it takes 3N of force to hold the mass in place. If the displacement from equilibrium is doubled, the force necessary to keep the system in place will __________.

**Possible Answers:**

increase by a factor of 2

increase by a factor of 4

remain the same

increase by a factor of

**Correct answer:**

increase by a factor of 2

Since the spring is Hookean, the relationship between the force and displacement from equilibrium of the mass can be expressed by Hooke's Law:

Since this equation is linear, the force and displacement are directly proportional. Thus, when the displacement doubles, the force doubles.

### Example Question #8 : Springs

Suppose that a 10kg object is sliding along a frictionless, horizontal surface at a speed of when suddenly, it hits and compresses a spring. If this spring has a spring constant of , what is the maximum length by which the spring is compressed?

**Possible Answers:**

**Correct answer:**

To find the answer to this problem, we'll need to consider the case before the object collides with the spring, and also after it collides. Before the collision, all of the object's energy is in the form of kinetic energy.

Upon collision, the spring will be compressed by the moving object. The maximum displacement, however, will occur the instant the object stops moving. At this point, the object will no longer have any kinetic energy because it has stopped moving. Furthermore, all of the energy has now been transferred into the spring, and is held as spring potential energy.

Also, since we are told in the question stem that no friction is involved, we have a situation in which mechanical energy is conserved. In other words, we can relate the initial and final energies as equal to one another.

Set these two equations equal to each other, and solve for the maximum displacement.

### Example Question #9 : Springs

What are the units of the spring constant in Hooke's Law?

**Possible Answers:**

**Correct answer:**

Hooke's Law is:

here is in meters and is in newtons.

Solve for to see what the units are:

This means that the units of the spring constant, is:

We can ignore the negative sign when we determine units, as the negative sign only indicates the direction of the force.

### Example Question #10 : Springs

A spring is compressed to its smallest point and then released. Assume that there is a frictional force that counters the motion of the spring. If we were to create a model plotting the position with time , which of the following could be appropriate for this system? Assume that at , is at its minimum point.

**Possible Answers:**

**Correct answer:**

To determine which type of sinusoid is used, we have to look at its starting point and where it goes. It starts at its minimum and moves towards its maxima. The simplest model we can use is a negative cosine model, which will require no phase shifts.

At , will be at its minimal value.

However, there is also a counteracting frictional force that decreases the amplitude of the spring over time. One way to do so would be to add an term in front of the cosine, so that we have an ever decreasing amplitude as time moves on.

The only appropriate model is

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