All AP Physics 1 Resources
Example Questions
Example Question #165 : Forces
If a spring of constant has a potential energy , calculate the compression of the spring.
From the definition of the spring potential,
Plugging in the given values lets us arrive at a calculated compression,
Example Question #30 : Spring Force
A vertical spring has a constant and an equilibrium height of . If you want to use the spring to launch an object of mass to a height of , how far must the spring be depressed from equilibrium?
We can calculate the energy needed to be stored in the spring from the change in potential energy that the object experiences:
Where our change in height is going to be the distance between its high point and the point when the spring is compressed. Furthermore, this change in potential energy will be equal to the energy stored in the spring:
was used since we're dealing with vertical distances. Setting these equal to each other:
We can also develop an expression for the change in height:
Substituting this back into our expression:
Distributing and rearranging:
Plugging in our values, we get:
Using the quadratic formula, we get:
or
Where a positive distance represents compression of the spring, and a negative distance represents extension of the spring. Therefore, the positive distance is our answer.
Example Question #161 : Forces
A spring is attached to the ceiling of an elevator with a block of mass hanging from it. If the displacement of the spring is while the elevator is at rest, what is the displacement of the spring when the elevator begins accelerating upward at a rate of .
When the elevator is at rest, we can use the following expression to determine the spring constant:
Where the force is simply the weight of the spring:
Rearranging for the constant:
Now solving for the constant:
Now applying the same equation for when the elevator is accelerating upward:
Where a is the acceleration due to gravity PLUS the acceleration of the elevator. Rearranging for the displacement:
Plugging in our values:
If you're confused why we added the acceleration of the elevator to the acceleration due to gravity. Think about the situation practically. When you are riding an elevator and it begins to accelerate upward, your body feels heavier. That's because your relative weight has increased due to the increased normal force due to a relative increase in acceleration.
Example Question #32 : Spring Force
A horizontal spring with a constant is sitting on a frictionless surface. If a force of is applied to the spring for and then a force of is applied for , how much work was done on the spring after ?
The important part of this problem is to not get bogged down in all of the unnecessary information. All we need to know to solve this problem is the spring constant and what force is being applied after 8s. Since the spring potential energy expression is a state function, what happens in between 0s and 8s is noncontributory to the question being asked. Therefore, we can determine the displacement of the spring using:
Rearranging for , we get:
As previously mentioned, we will be using the force that is being applied at :
Then using the expression for potential energy of a spring:
Where potential energy is the work we are looking for. Plugging in our values:
Example Question #33 : Spring Force
A horizontal spring with constant is on a frictionless surface with an attached block of mass . How much force must initially be applied to the block so that its maximum velocity is ? Assume simple harmonic motion.
First, let's begin with the force expression for a spring:
Rearranging for displacement, we get:
Then we can substitute this into the expression for potential energy of a spring:
We should note that this is the maximum potential energy the spring will achieve. Also, we know that the maximum potential energy of a spring is equal to the maximum kinetic energy of a spring:
Therefore:
Substituting in the expression for kinetic energy:
Now rearranging for force, we get:
We have all of these values, so we can solve the problem:
Example Question #31 : Spring Force
A spring with constant is at equilibrium and hanging vertically from a ceiling. If a block of mass is attached to the spring and pulled down , what is the instantaneous acceleration of the block when it is released?
We can use Newton's second law to solve this problem:
There are two forces acting on the block, the force of gravity and the force from the spring. If we designate an upward force as being positive, we can then say:
Rearranging for acceleration, we get:
Plugging in our values, we get:
Therefore, the block is already at equilibrium and will not move upon being released.
Example Question #35 : Spring Force
A horizontal spring with constant is on a frictionless surface with a block attached to one end. If the spring is compressed and the instantaneous acceleration of the block is after being released, what is the mass of the block?
We can use Newton's second law to solve this problem:
Where the only force is from the spring, so we can say:
Rearranging for mass, we get:
Plugging in our values, we get:
Example Question #36 : Spring Force
A horizontal spring with constant is on a surface with . A block of mass is attached to the end of the spring. If the spring is compressed by and released, what is the velocity of the block as it passes through the equilibrium of the spring?
We can use the expression for conservation of energy to solve this problem:
There is no initial kinetic (starts at rest) or final potential (at equilibrium), so we can say:
Where work is done by friction. Inserting expressions for each of these, we get:
Multiplying both sides of the equation by 2 and rearranging for velocity, we get:
Plugging in values for each of these variables, we get:
Example Question #31 : Spring Force
A spring is used to swing a mass at . If the spring stretches by , determine the spring constant.
None of these
The radius of the circle will be . This is the rest length plus the stretch of the spring.
Thus, the circumference will be
Since the angular velocity is
Thus, the linear velocity is
The force of the spring will be equal to the centripetal force.
Solving for the spring constant:
Plugging in values:
Example Question #31 : Spring Force
If a board depresses identical parallel springs by . Determine the spring constant.
Using
Solving for
Converting to and plugging in values: