Work, Energy, and Power - MCAT Chemical and Physical Foundations of Biological Systems
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An empty mining cart has a mass of 
and is traveling down a track that has a slope of 
to the horizontal. The cart is traveling at a rate of 
when an operator notices a disturbance on the track ahead and locks the wheels of the cart. What is the speed of the cart after it has traveled 
with the wheels locked?
An empty mining cart has a mass of
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We need the equation for conservation of energy for this problem:
We can eliminate final potential energy if we set the final height to be zer. We are solving for final velocity, so let's rearrange for final kinetic energy.
Substituting our equations for each variable, we get:
Rearranging for final velocity we get:
If you derive this formula and are unsure of your work, simply check your units. Each term under the square root has units of 
, which will ultimately give us units of 
, which is what we want.
We have values for all variables except two: height and normal force.
Let's calculate the height. We know that the between the initial and final states, the cart has traveled 20 meters down a slope of 40 degrees. Therefore, we can calculate height with the formula:
Now we just need to find the normal force. The following diagram will help visualize this calculation.
If you are unsure whether to use sine or cosine, think about it practically. As the angle gets less and less, the normal force is going to get larger. This is characteristic of a cosine function.
Therefore, we can say that:
Now that we have all of our variables, it's time to plug and chug:
We need the equation for conservation of energy for this problem:
We can eliminate final potential energy if we set the final height to be zer. We are solving for final velocity, so let's rearrange for final kinetic energy.
Substituting our equations for each variable, we get:
Rearranging for final velocity we get:
If you derive this formula and are unsure of your work, simply check your units. Each term under the square root has units of
We have values for all variables except two: height and normal force.
Let's calculate the height. We know that the between the initial and final states, the cart has traveled 20 meters down a slope of 40 degrees. Therefore, we can calculate height with the formula:
Now we just need to find the normal force. The following diagram will help visualize this calculation.
If you are unsure whether to use sine or cosine, think about it practically. As the angle gets less and less, the normal force is going to get larger. This is characteristic of a cosine function.
Therefore, we can say that:
Now that we have all of our variables, it's time to plug and chug:
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?
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?
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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
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
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?
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?
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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
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
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.
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?
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.
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?
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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.
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.
A stone of mass m sits atop a hill of height h. As it rolls downhill, which of the following is true?
- Half way down the hill, PE=\frac{1}{2}$mgh
- Half way down the hill, PE=\frac{1}{4}$mgh
- Half way down the hill, PE=mg$\sqrt{h}$
- Half way down the hill, PE still equals mgh.
- None of these is true.
A stone of mass m sits atop a hill of height h. As it rolls downhill, which of the following is true?
- Half way down the hill, PE=\frac{1}{2}$mgh
- Half way down the hill, PE=\frac{1}{4}$mgh
- Half way down the hill, PE=mg$\sqrt{h}$
- Half way down the hill, PE still equals mgh.
- None of these is true.
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4. Choice 1 is correct because initially, all of the mechanical energy in the stone was potential energy and none was kinetic energy: ME = KE + PE. PE is “stored” in the stone-hill system by rolling the stone up hill. It is obvious that it takes half as much energy to roll the stone half way up the hill, compared with rolling it to the top. At the bottom of the hill, all of the PE will have been converted into KE, given by the formula $KE=\frac{1}{2}$mv^{2}$
Since the PE was ½ mgh when rolling the stone half way up hill, it is the same as it rolls down hill.
4. Choice 1 is correct because initially, all of the mechanical energy in the stone was potential energy and none was kinetic energy: ME = KE + PE. PE is “stored” in the stone-hill system by rolling the stone up hill. It is obvious that it takes half as much energy to roll the stone half way up the hill, compared with rolling it to the top. At the bottom of the hill, all of the PE will have been converted into KE, given by the formula $KE=\frac{1}{2}$mv^{2}$
Since the PE was ½ mgh when rolling the stone half way up hill, it is the same as it rolls down hill.
A ball with mass of 2kg is dropped from the top of a building this is 30m high. What is the approximate velocity of the ball when it is 10m above the ground?
A ball with mass of 2kg is dropped from the top of a building this is 30m high. What is the approximate velocity of the ball when it is 10m above the ground?
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Use conservation of energy. The gravitational potential energy lost as the ball drops from 30m to 10m equals the kinetic energy gained.
Change in gravitational potential energy can be found using the difference in mgh. 
So 400 Joules are converted from gravitational potential to kinetic energy, allowing us to solve for the velocity, v.
400 Joules = $$\frac{1}{2}$mv^{2}$
400 = $v^{2}$
v = 20 m/s
Use conservation of energy. The gravitational potential energy lost as the ball drops from 30m to 10m equals the kinetic energy gained.
Change in gravitational potential energy can be found using the difference in mgh.
So 400 Joules are converted from gravitational potential to kinetic energy, allowing us to solve for the velocity, v.
400 Joules = $$\frac{1}{2}$mv^{2}$
400 = $v^{2}$
v = 20 m/s
Consider a spring undergoing simple harmonic motion. When the spring is at its maximum velocity .
Consider a spring undergoing simple harmonic motion. When the spring is at its maximum velocity .
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Kinetic energy is highest when the spring is moving the fastest. Conversely, potential energy is highest when the spring is most compressed, and momentarily stationary. When the force resulting from the compression causes the spring to extend, potential energy decreases as velocity increases.
Kinetic energy is highest when the spring is moving the fastest. Conversely, potential energy is highest when the spring is most compressed, and momentarily stationary. When the force resulting from the compression causes the spring to extend, potential energy decreases as velocity increases.
A pendulum with a mass of 405kg reaches a maximum height of 2.4m. What is its velocity at the bottommost point in its path?
A pendulum with a mass of 405kg reaches a maximum height of 2.4m. What is its velocity at the bottommost point in its path?
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First solve for the potential energy of the pendulum at the height of 2.4m.
PE = mgh
PE = (405kg)(10m/s2)(2.4m) = 9720J
This must be equal to the maximum kinetic energy of the object.
KE = ½mv2
9720J = ½mv2
Plug in the mass of the object (405 kg) and solve for v.
9720J = ½(405kg)v2
v = 6.9m/s
First solve for the potential energy of the pendulum at the height of 2.4m.
PE = mgh
PE = (405kg)(10m/s2)(2.4m) = 9720J
This must be equal to the maximum kinetic energy of the object.
KE = ½mv2
9720J = ½mv2
Plug in the mass of the object (405 kg) and solve for v.
9720J = ½(405kg)v2
v = 6.9m/s
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.
At the bottom of a neighboring hill, a neighbor watches Sally and Sam come down the hill. Sally is traveling 15m/s and Sam is traveling 10m/s. From the moment the neighbor begins watching, to just after they both come to a stop, who has dissipated more heat in the form of friction? (Assume all friction is lost as heat).
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.
At the bottom of a neighboring hill, a neighbor watches Sally and Sam come down the hill. Sally is traveling 15m/s and Sam is traveling 10m/s. From the moment the neighbor begins watching, to just after they both come to a stop, who has dissipated more heat in the form of friction? (Assume all friction is lost as heat).
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Sally has greater kinetic energy in this example than does Sam. From the moment when the neighbor begins watching we can calculate the kinetic energy. Once stopped, all of the kinetic energy will have been dissipated.
Sally's KE = 1/2 (52kg) (15m/s)2 = 5850J
Sam's KE = 1/2 (60kg) (10m/s)2 = 3000J
All of this energy will be dissipated as friction before Sam and Sally come to a stop.
Sally has greater kinetic energy in this example than does Sam. From the moment when the neighbor begins watching we can calculate the kinetic energy. Once stopped, all of the kinetic energy will have been dissipated.
Sally's KE = 1/2 (52kg) (15m/s)2 = 5850J
Sam's KE = 1/2 (60kg) (10m/s)2 = 3000J
All of this energy will be dissipated as friction before Sam and Sally come to a stop.
A rock is dropped from a given height and allowed to hit the ground. The velocity of the rock is measured upon impact with the ground. Assume there is no air resistance.
In order for the velocity of the rock to be doubled before impact, which of the following is necessary?
A rock is dropped from a given height and allowed to hit the ground. The velocity of the rock is measured upon impact with the ground. Assume there is no air resistance.
In order for the velocity of the rock to be doubled before impact, which of the following is necessary?
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We can compare height and velocity by comparing the equations for potential and kinetic energy. This is possible because the rock initially has no kinetic energy (velocity is zero) and has no potential energy upon impact (height is zero). Using conservation of energy will yield the comparison below.
Because velocity is squared in the equation for kinetic energy, it requires a quadrupling of height in order to double the velocity.
We can compare height and velocity by comparing the equations for potential and kinetic energy. This is possible because the rock initially has no kinetic energy (velocity is zero) and has no potential energy upon impact (height is zero). Using conservation of energy will yield the comparison below.
Because velocity is squared in the equation for kinetic energy, it requires a quadrupling of height in order to double the velocity.
A 
boulder drops from rest off of a 
cliff. Find its velocity at 
before impact.
A
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Conservation of energy dictates that the initial energy and final energy will be equal.
In this case, the boulder starts with zero kinetic energy and ends with both kinetic and potential energy.
We can cancel the mass from each term and plug in the given values to solve for the velocity at a height of 
.
Conservation of energy dictates that the initial energy and final energy will be equal.
In this case, the boulder starts with zero kinetic energy and ends with both kinetic and potential energy.
We can cancel the mass from each term and plug in the given values to solve for the velocity at a height of
Ignoring air resistance, which of the following is true regarding the motion of a pendulum?
Ignoring air resistance, which of the following is true regarding the motion of a pendulum?
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Energy must be conserved through the motion of a pendulum. Let point 1 represent the bottom of the oscillation and point 2 represent the top. At point 1, there is no potential energy, using point 1 as our "ground/reference," thus all of the system energy is kinetic energy. At point 2, the velocity is zero; thus, the kinetic energy is zero and all of the system energy is potential energy. At the highest point in the swing, potential energy is at a maximum, and at the lowest point in the swing, kinetic energy is at a maximum.
Energy must be conserved through the motion of a pendulum. Let point 1 represent the bottom of the oscillation and point 2 represent the top. At point 1, there is no potential energy, using point 1 as our "ground/reference," thus all of the system energy is kinetic energy. At point 2, the velocity is zero; thus, the kinetic energy is zero and all of the system energy is potential energy. At the highest point in the swing, potential energy is at a maximum, and at the lowest point in the swing, kinetic energy is at a maximum.
A block of wood is floating in space. A bullet is fired from a gun and hits the block, embedding itself in the wood and generating heat. Which of the following is conserved?
A block of wood is floating in space. A bullet is fired from a gun and hits the block, embedding itself in the wood and generating heat. Which of the following is conserved?
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Momentum is always conserved in a system, when not experiencing external forces.
An inelastic collision can be identified if the two objects stick together after the collision occurs, such as the bullet becoming embedded in the wood. In an inelastic collision, kinetic energy is not conserved. Since mechanical energy is the sum of kinetic and potential energy, mechanical energy is also not conserved. This lack of conservation is due to the conversion of some of the kinetic energy to heat and sound. The kinetic energy decreases as the heat energy increases, resulting in a non-constant temperature.
Momentum is always conserved in a system, when not experiencing external forces.
An inelastic collision can be identified if the two objects stick together after the collision occurs, such as the bullet becoming embedded in the wood. In an inelastic collision, kinetic energy is not conserved. Since mechanical energy is the sum of kinetic and potential energy, mechanical energy is also not conserved. This lack of conservation is due to the conversion of some of the kinetic energy to heat and sound. The kinetic energy decreases as the heat energy increases, resulting in a non-constant temperature.
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.
If the hill is frictionless, and after both Sally and Sam reach the bottom of the hill traveling at maximum velocity, who has lost more potential energy?
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.
If the hill is frictionless, and after both Sally and Sam reach the bottom of the hill traveling at maximum velocity, who has lost more potential energy?
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When they reach the bottom of the hill, Sam and Sally have both converted all of their potential energy into kinetic energy. We measured potential energy at 50m, so once they have lost 50m, the potential energy is 0, while the kinetic energy has reached maximum value. Because Sam has greater mass, he had more potential energy to convert to kinetic energy.
When they reach the bottom of the hill, Sam and Sally have both converted all of their potential energy into kinetic energy. We measured potential energy at 50m, so once they have lost 50m, the potential energy is 0, while the kinetic energy has reached maximum value. Because Sam has greater mass, he had more potential energy to convert to kinetic energy.
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.
After reaching maximum velocity at the bottom of a frictionless hill identical to the one in the question, Sam deploys a parachute. How much energy must the parachute dissipate before Sam comes to a stop?
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.
After reaching maximum velocity at the bottom of a frictionless hill identical to the one in the question, Sam deploys a parachute. How much energy must the parachute dissipate before Sam comes to a stop?
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The parachute must dissipate all of the kinetic energy that Sam has at the bottom of the hill, since there is no friction to slow him. He has 30kJ of energy, as per the equation PE = mgh.
PE = 60kg * 10m/s2 * 50m = 30,000J = KE at the bottom of the hill
The parachute must dissipate all of the kinetic energy that Sam has at the bottom of the hill, since there is no friction to slow him. He has 30kJ of energy, as per the equation PE = mgh.
PE = 60kg * 10m/s2 * 50m = 30,000J = KE at the bottom of the hill
Two children are playing on an icy lake. Child 1 weighs 50kg, and child 2 weighs 38kg. Child 1 has a backpack that weighs 10kg, and child 2 has a backpack that weighs 5kg.
Over the course of the afternoon, they collide many times. Four collisions are described below.
Collision 1:
Child 1 starts from the top of a ramp, and after going down, reaches the lake surface while going 
and subsequently slides into a stationary child 2. They remain linked together after the collision.
Collision 2:
Child 1 and child 2 are sliding in the same direction. Child 2, moving at 
, slides into child 1, moving at 
.
Collision 3:
The two children collide while traveling in opposite directions at 
each.
Collision 4:
The two children push off from one another’s back, and begin moving in exactly opposite directions. Child 2 moves with a velocity of 
.
In collision 1, imagine that child 2 was not present on the ice. How much energy would child 1 have to dissipate to the lake surface before he came to a stop? Ignore wind resistance.
Two children are playing on an icy lake. Child 1 weighs 50kg, and child 2 weighs 38kg. Child 1 has a backpack that weighs 10kg, and child 2 has a backpack that weighs 5kg.
Over the course of the afternoon, they collide many times. Four collisions are described below.
Collision 1:
Child 1 starts from the top of a ramp, and after going down, reaches the lake surface while going
Collision 2:
Child 1 and child 2 are sliding in the same direction. Child 2, moving at
Collision 3:
The two children collide while traveling in opposite directions at
Collision 4:
The two children push off from one another’s back, and begin moving in exactly opposite directions. Child 2 moves with a velocity of
In collision 1, imagine that child 2 was not present on the ice. How much energy would child 1 have to dissipate to the lake surface before he came to a stop? Ignore wind resistance.
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The amount of kinetic energy that child 1 has after she reaches the lake surface is the amount of energy she will dissipate to the lake before coming to a complete stop.
The amount of kinetic energy that child 1 has after she reaches the lake surface is the amount of energy she will dissipate to the lake before coming to a complete stop.
A 5kg ball is attached to a 10m rope. The ball is held at a horizontal angle and allowed to fall freely with a pendulum-like motion. Assume there is no air resistance.
What is the velocity of the ball when it is at its lowest point?
A 5kg ball is attached to a 10m rope. The ball is held at a horizontal angle and allowed to fall freely with a pendulum-like motion. Assume there is no air resistance.
What is the velocity of the ball when it is at its lowest point?
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This problem becomes much easier if we use the law of conservation of energy. Because there is no friction involved, all of the ball's potential energy was converted into kinetic energy when falling through the pendulum arc. As a result, the velocity at the bottom of the swing can be determined using the equation below.
Because the potential energy at the bottom of the swing is zero (relative to the starting point), and the velocity at the top is zero, we can simplify the equation.
Since the string is 10m long, we know that the lowest point will be 10m below the starting point.
This problem becomes much easier if we use the law of conservation of energy. Because there is no friction involved, all of the ball's potential energy was converted into kinetic energy when falling through the pendulum arc. As a result, the velocity at the bottom of the swing can be determined using the equation below.
Because the potential energy at the bottom of the swing is zero (relative to the starting point), and the velocity at the top is zero, we can simplify the equation.
Since the string is 10m long, we know that the lowest point will be 10m below the starting point.
How much energy is required to a accelerate a 
block from rest to a final speed of 
?
How much energy is required to a accelerate a
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From an energy stand-point, the block starts with zero kinetic energy and zero potential energy. At the end, the block still has zero potential energy, but has a non-zero kinetic energy. Assuming there is no friction, all energy added to the block have been converted to kinetic energy.
Using the change in velocity, we can solve for the energy used to move the block.
We are given our mass and the change in velocity, allowing us to solve for the change in kinetic energy.
From an energy stand-point, the block starts with zero kinetic energy and zero potential energy. At the end, the block still has zero potential energy, but has a non-zero kinetic energy. Assuming there is no friction, all energy added to the block have been converted to kinetic energy.
Using the change in velocity, we can solve for the energy used to move the block.
We are given our mass and the change in velocity, allowing us to solve for the change in kinetic energy.
What is the kinetic energy of a 
bullet moving at 
?
What is the kinetic energy of a
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Kinetic energy is given by the formula:
We are given the mass of the bullet and its velocity, allowing us to calculate kinetic energy from this formula.
Work is often expressed in Newton-meters, but energy is usually expressed as Joules, although the units are equivalent.
Kinetic energy is given by the formula:
We are given the mass of the bullet and its velocity, allowing us to calculate kinetic energy from this formula.
Work is often expressed in Newton-meters, but energy is usually expressed as Joules, although the units are equivalent.
What is the kinetic energy of a 0.1kg projectile traveling at 
?
What is the kinetic energy of a 0.1kg projectile traveling at
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We can calculate the kinetic energy using the equation:
Use the given values for the mass and velocity to solve:
We can calculate the kinetic energy using the equation:
Use the given values for the mass and velocity to solve: