All High School Physics Resources
Example Questions
Example Question #1 : Energy Of Shm
A mass is placed at the end of a spring. The spring is compressed . What is the maximum velocity of the mass if the spring has a spring constant of ?
If we're looking for the maximum velocity, that will happen when all the energy in the system is kinetic energy.
We can use the law of conservation of energy to see . So, if we can find the initial potential energy, we can find the final kinetic energy, and use that to find the mass's final velocity.
The formula for spring potential energy is:
Plug in our given values and solve:
The formula for kinetic energy is:
Since , that means that .
We can plug in that information to the formula for kinetic energy to solve for the maximum velocity:
Example Question #2 : Energy Of Shm
A mass on a string is released and swings freely. Which of the following best explains the energy of the pendulum when the string is perpendicular to the ground?
The mass has maximum kinetic energy
The mass has equal amounts of kinetic and potential energy
The mass has maximum potential energy
The mass has mostly kinetic energy, but there is some potential energy
The mass has mostly potential energy, but there is some kinetic energy
The mass has maximum kinetic energy
Conservation of energy dictates that the total mechanical energy will remain constant. Initially, the mass will not be moving and will be at its highest height. When released, it will begin to travel downward (lose potential energy) and gain velocity (gain kinetic energy). When the mass reaches the bottommost point in the swing, the potential energy will be at a minimum and the kinetic energy will be at a maximum. This point corresponds to the string being perpendicular to the ground.
Example Question #2 : Energy Of Shm
Derive a formula for the maximum speed of a simple pendulum bob in terms of , the length L and the angle of swing .
We can start out by examining the energy in the pendulum. At the top of the swing, there is gravitational potential energy. At the bottom of the swing there is kinetic energy. The law of conservation of energy states that these two values must be the same.
We know the equations for gravitational potential energy and kinetic energy to be
We can set these equal to each other.
Since the mass is constant, it falls out of both sides of the equation.
Let’s rearrange and solve for v by itself.
We don’t know the height of the pendulum at this point and need to get it in terms of the length of the pendulum.
If we knew the angle that the pendulum made with the vertical equilibrium point, we could determine how far off the ground the pendulum was. We can create a triangle with the hypotenuse as the length of the string and the angle between the pendulum and the equilibrium point.
This adjacent side can then be subtracted from the original length of the pendulum to determine the height off the ground.
We can substitute this back into our equation
Example Question #3 : Energy Of Shm
A bullet with mass hits a ballistic pendulum with length and mass and lodges in it. When the bullet hits the pendulum it swings up from the equilibrium position and reaches an angle at its maximum. Determine the bullet’s velocity.
We will need to start at the end of the situation and work backwards in order to determine the velocity of the bullet. At the very end, the pendulum with the bullet reaches its maximum height and therefore comes to a stop. It has gravitational potential energy. At the bottom of the pendulum right after the bullet collides with it, it has kinetic energy due to the velocity of the bullet. With the law of conservation of energy we can set the kinetic energy of the pendulum right after the collision equal to the gravitational potential energy of the pendulum at the highest point.
To determine the height of the pendulum we will need to use trig and triangles to find the height. We know that the pendulum makes a 30 degree angle with the equilibrium position at its maximum height. The length of the pendulum is provided which is the hypotenuse of this triangle. We need to find the adjacent side of this triangle. We can use cosine to determine this.
We can now subtract this value from the length of the pendulum to determine how high off the ground the pendulum is at its highest point.
We can now set the kinetic energy of the pendulum right after the collision equal to the gravitational potential energy of the pendulum at the highest point.
The mass is the same throughout so it falls out of the equation.
The pendulum with the bullet was moving after the collision. We can now use momentum to determine the speed of the bullet before the collision. Conservation of momentum states that the momentum before the collision must equal the momentum after the collision.
They both move together after the collision
Since the pendulum was not moving at the beginning
We can now plug in these values and solve for the missing piece.
Example Question #4 : Energy Of Shm
A spring with a spring constant of is compressed . How much potential energy has been generated?
The formula for the potential energy in a spring is:
Use the given spring constant and displacement to solve for the stored energy.
Example Question #3 : Energy Of Shm
A pendulum is dropped from rest at above the ground. If no outside forces (except gravity) are acting upon it, what will be its maximum height on the other side?
Less than
Greater than
If no outside forces act upon the pendulum, it will continue to oscillate back to the original height of .
The proof of this is in the law of conservation of energy. At the top, the pendulum has all potential energy, which is given by the formula . As it swings, the potential energy is converted to kinetic energy until, at the bottommost point, there is only kinetic energy. It then changes direction and begins to rise again. When it rises to the maximum height on the other side, all of its kinetic energy will turn back into potential energy.
Mathematically, the initial and final potential energies are equal.
Notice the masses and gravity can cancel out on both sides, as neither of these will change. This leaves us with only height.
Example Question #1 : Circular Motion
A shop sign weighing hangs from the end of a uniform beam as shown. Find the horizontal and vertical forces exerted by the hinge on the beam at the wall.
This is a static equilibrium problem. In order for static equilibrium to be achieved, there are three things that must be true. First, the sum of the forces in the horizontal direction must all equal . Second, the sum of the forces in the vertical direction must all be equal to zero. Third, the torque around a fixed axis must equal .
Let us begin by summing up the forces in the vertical direction.
Then let us sum up the forces in the horizontal direction
Lastly let us analyze the torque, using the hinge as the axis point where is the length of the beam.
Looking at these three equations, the easiest to work with would be our torque equation as it is only missing one variable. If we are able to find the tension in the direction we would then be able to use trigonometry to determine the tension in the cable overall.
We can now substitute this value back into our vertical direction equation to determine the force on the hinge in the direction.
We can go back to our tension in our direction and use trigonometry to determine the force of the tension in direction.
Rearrange and solve for the tension in the direction.
We can now go back to our second equation and substitute this value in.
Example Question #2 : Circular Motion
A pulley has a bucket of weight hanging from the cord over a well. The pulley has a mass of and radius of . There is a frictional torque of at the axle. Assume the cord has negligible mass and does not stretch or slip on the pulley. Calculate the linear acceleration of the bucket.
First let us analyze the torque that is happening on the bucket. There is a torque from the friction on the pulley and there is a torque from the bucket pulling on the pulley.
Next let us analyze the forces involved. There is a tension force pulling up on the bucket and there is the force of gravity pulling down on the bucket.
We can rearrange this to find so that we can substitute it into our torque equation.
Now substitute this into our torque equation
We know that there is a relationship between acceleration and angular acceleration.
So we can substitute this into our equation so that only linear acceleration is present.
We also know that the moment of inertia of the system is equal to the moment of inertia of the pulley plus the bucket.
We can now substitute this into our equation.
We can now starting putting in our known variables to solve for the missing acceleration.
Example Question #3 : Circular Motion
The torque applied to a wrench is . If the force applied to the wrench is , how long is the wrench?
The formula for torque is:
We are given the total torque and the force applied. Using these values, we can solve for the length of the wrench.
Example Question #4 : Circular Motion
Two children are trying to balance on a see-saw. One child has a mass of the other has a mass of . If the see-saw is balanced perfectly in the middle and the child is sitting at one end of the see-saw, how far from the center should the child sit so that the system is perfectly balanced?
If the see-saw is in total, then it has on either side of the fulcrum.
The question is asking us to find the equilibrium point; that means we want the net torque to equal zero.
Now, find the torque for the first child.
We are going to use the force of gravity for the force of the child.
When thinking of torque, treat the positive/negative as being clockwise vs. counter-clockwise instead of up vs. down. In this case, child one is generating counter-clockwise torque. That means that since , child two will be generating clockwise torque.
Solve for the radius (distance) of the second child.
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