Physics 6A Work and Energy examples Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.

Physics 6A Work and Energy examples Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Work and Energy Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Energy comes in many forms. We will most often encounter two kinds of energy: Kinetic Energy – Energy of Motion. Any moving object has kinetic energy. The formula is KE = ½ mv 2 Potential Energy – Stored Energy. There will be several types of potential energy: * Gravitational – Energy stored by lifting an object above the earth. We will have a more robust formula later, but for now: U grav = mgh * Elastic – Energy stored by stretching or compressing a spring. The formula is U elastic = ½ kx 2 * Electric – Energy stored by charges in an electric field. We will see this next quarter.

Work and Energy Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Work is energy transferred to a moving object when a force acts on it. To do work, the force must line up with the motion of the object. Perpendicular forces do no work. We will have two formulas involving work. W = Fdcos(θ) W = ΔKE Our main concept that ties it all together is Conservation of Energy. This says that the total energy of a system does not change. We can write down a formula that accounts for all the forms of energy: KE i + U i + W NC = KE f + U f This will be the template for most of the problems you will do involving energy.

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: A 100kg box (initially at rest) is pushed across a horizontal floor by a force of 400N. If the coefficients of friction are μ k =0.2 and μ s =0.4, find the total work done on the box and the final speed when the box is pushed 10m. Assume the applied force is horizontal.

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: A 100kg box (initially at rest) is pushed across a horizontal floor by a force of 400N. If the coefficients of friction are μ k =0.2 and μ s =0.4, find the total work done on the box and the final speed when the box is pushed 10m. Assume the applied force is horizontal. 400 N weight Normal force friction Here is the free-body diagram. Since the box is moving horizontally, the only forces that do work are friction and the 400N push. Calculate the force of kinetic friction: Work done by each force: Total work done on box: 10m

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: A 100kg box on a horizontal floor (initially at rest) is pushed by a force of 400N, applied downward at an angle of 30°. If the coefficients of friction are μ k =0.2 and μ s =0.4, find the total work done on the box. Does the box move?

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: A 100kg box on a horizontal floor (initially at rest) is pushed by a force of 400N, applied downward at an angle of 30°. If the coefficients of friction are μ k =0.2 and μ s =0.4, find the total work done on the box. Does the box move? 400 N weight Normal force friction 30° In the last question, we assumed the box was moving because the problem told us ho far it moved. This one is different, and we have to determine whether or not the box even moves. To do this, we should find the maximum friction force and compare to the forward push – if the push is not enough to overcome static friction the box will not move. First we will need to break the 400N push into components. The downward component will increase the normal force.

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: A 100kg box on a horizontal floor (initially at rest) is pushed by a force of 400N, applied downward at an angle of 30°. If the coefficients of friction are μ k =0.2 and μ s =0.4, find the total work done on the box. Does the box move? 400 N weight Normal force friction 30° First we will need to break the 400N push into components. 200 N 346 N To get the max friction we need the normal force, which will be greater because of the downward push: Compare this to the forward component of the push (only 346 N). Friction is strong enough to hold the box in place, so there is no motion. No work is done on the box. Note that the actual friction force is only 346N – just enough to hold the box.

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: A 100kg box is released from rest at the top of a frictionless 10-meter high ramp that makes an angle of 30° with the horizontal. Find the final speed of the box when it reaches the bottom of the ramp. Compare to the impact speed when the box is pushed over the edge and free- falls to the ground instead. 10m 30°

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: A 100kg box is released from rest at the top of a frictionless 10-meter high ramp that makes an angle of 30° with the horizontal. Find the final speed of the box when it reaches the bottom of the ramp. Compare to the impact speed when the box is pushed over the edge and free- falls to the ground instead. 10m 30° We can do this one with conservation of energy. Here is the basic format: KE i + U i + W NC = KE f + U f 0 + mgh + 0 = ½ mv 2 + 0 Solving for v: If the box is pushed over the edge, we can use conservation of energy again, and get the exact same result. This happens because there was no friction on the ramp, so the speed at the bottom is the same in both cases.

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: A 100kg box is released from rest at the top of a 10-meter high ramp that makes an angle of 30° with the horizontal. Assume the coefficients of friction are μ k =0.2 and μ s =0.3. Find the final speed of the box when it reaches the bottom of the ramp. 10m 30°

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: A 100kg box is released from rest at the top of a 10-meter high ramp that makes an angle of 30° with the horizontal. Assume the coefficients of friction are μ k =0.2 and μ s =0.3. Find the final speed of the box when it reaches the bottom of the ramp. 10m 30° We can do this one with conservation of energy. Here is the basic format: KE i + U i + W NC = KE f + U f 0 + mgh + W friction = ½ mv 2 + 0 We need to find the work done by kinetic friction as the box slides down the ramp. Also, does it even slide, or is there enough static friction to hold it in place?

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: A 100kg box is released from rest at the top of a 10-meter high ramp that makes an angle of 30° with the horizontal. Assume the coefficients of friction are μ k =0.2 and μ s =0.3. Find the final speed of the box when it reaches the bottom of the ramp. 30° Does the box slide at all? We should draw the free-body diagram to see what forces are in play. The friction force is related to the Normal force. mg Normal force friction mgcos30 mgsin30 Let’s calculate the max static friction and the downhill gravity force to see which is bigger. Looks like plenty of downhill force to overcome friction. The actual friction will be kinetic: As the box slides, kinetic friction will do work against the motion.

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: A 100kg box is released from rest at the top of a 10-meter high ramp that makes an angle of 30° with the horizontal. Assume the coefficients of friction are μ k =0.2 and μ s =0.3. Find the final speed of the box when it reaches the bottom of the ramp. 10m 30° Now we can fill in the conservation of energy formula. 0 + mgh + W friction = ½ mv 2 + 0 Work done by kinetic friction: d

1) A boy exerts a force of 11.0N at 29.0 degrees above the horizontal on a 6.40kg sled. Find the work done by the boy and the final speed of the sled after it moves 2.00m, assuming the sled starts with an initial speed of 0.500m/s and slides horizontally without friction. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

1) A boy exerts a force of 11.0N at 29.0 degrees above the horizontal on a 6.40kg sled. Find the work done by the boy and the final speed of the sled after it moves 2.00m, assuming the sled starts with an initial speed of 0.500m/s and slides horizontally without friction. 29° F=11N Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

1) A boy exerts a force of 11.0N at 29.0 degrees above the horizontal on a 6.40kg sled. Find the work done by the boy and the final speed of the sled after it moves 2.00m, assuming the sled starts with an initial speed of 0.500m/s and slides horizontally without friction. 29° F=11N Initially the sled is moving at 0.5 m/s, so its kinetic energy is: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

1) A boy exerts a force of 11.0N at 29.0 degrees above the horizontal on a 6.40kg sled. Find the work done by the boy and the final speed of the sled after it moves 2.00m, assuming the sled starts with an initial speed of 0.500m/s and slides horizontally without friction. 29° F=11N Initially the sled is moving at 0.5 m/s, so its kinetic energy is: Next we can find the work done by the boy’s pull, and add that to the kinetic energy. Remember – work always equals the change in the kinetic energy. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

1) A boy exerts a force of 11.0N at 29.0 degrees above the horizontal on a 6.40kg sled. Find the work done by the boy and the final speed of the sled after it moves 2.00m, assuming the sled starts with an initial speed of 0.500m/s and slides horizontally without friction. 29° F=11N The force is not aligned with the motion, so we need to use the x-component to get the work. Initially the sled is moving at 0.5 m/s, so its kinetic energy is: Next we can find the work done by the boy’s pull, and add that to the kinetic energy. Remember – work always equals the change in the kinetic energy. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

1) A boy exerts a force of 11.0N at 29.0 degrees above the horizontal on a 6.40kg sled. Find the work done by the boy and the final speed of the sled after it moves 2.00m, assuming the sled starts with an initial speed of 0.500m/s and slides horizontally without friction. 29° F=11N The force is not aligned with the motion, so we need to use the x-component to get the work. Initially the sled is moving at 0.5 m/s, so its kinetic energy is: Next we can find the work done by the boy’s pull, and add that to the kinetic energy. Remember – work always equals the change in the kinetic energy. Now the total KE is 20.04J. Use this to solve for the new speed: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

2) A spring-loaded toy gun is used to shoot a ball of mass m=1.50kg straight up in the air, as shown in the figure. The spring has a spring constant of k=667N/m. If the spring is compressed a distance of 25.0cm from its equilibrium position y=0 and then released, find the ball’s maximum height h max (measured from the equilibrium position of the spring.) There is no air resistance, and the ball never touches the inside of the gun. Assume that all movement occurs in a straight line up and down along the y-axis. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

2) A spring-loaded toy gun is used to shoot a ball of mass m=1.50kg straight up in the air, as shown in the figure. The spring has a spring constant of k=667N/m. If the spring is compressed a distance of 25.0cm from its equilibrium position y=0 and then released, find the ball’s maximum height h max (measured from the equilibrium position of the spring.) There is no air resistance, and the ball never touches the inside of the gun. Assume that all movement occurs in a straight line up and down along the y-axis. We will use conservation of energy for this one. Notice that the picture already has y=0 defined for us at the top of the tube (this is also the equilibrium position of the spring). [1] [3] Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

2) A spring-loaded toy gun is used to shoot a ball of mass m=1.50kg straight up in the air, as shown in the figure. The spring has a spring constant of k=667N/m. If the spring is compressed a distance of 25.0cm from its equilibrium position y=0 and then released, find the ball’s maximum height h max (measured from the equilibrium position of the spring.) There is no air resistance, and the ball never touches the inside of the gun. Assume that all movement occurs in a straight line up and down along the y-axis. We will use conservation of energy for this one. Notice that the picture already has y=0 defined for us at the top of the tube (this is also the equilibrium position of the spring). At the beginning (I will call this position [1]) the energy of the system is all potential – the ball is not moving so K=0. We need to account for the compression of the spring, as well as the gravitational potential energy of the ball: [1] [3] Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

2) A spring-loaded toy gun is used to shoot a ball of mass m=1.50kg straight up in the air, as shown in the figure. The spring has a spring constant of k=667N/m. If the spring is compressed a distance of 25.0cm from its equilibrium position y=0 and then released, find the ball’s maximum height h max (measured from the equilibrium position of the spring.) There is no air resistance, and the ball never touches the inside of the gun. Assume that all movement occurs in a straight line up and down along the y-axis. We will use conservation of energy for this one. Notice that the picture already has y=0 defined for us at the top of the tube (this is also the equilibrium position of the spring). At the beginning (I will call this position [1]) the energy of the system is all potential – the ball is not moving so K=0. We need to account for the compression of the spring, as well as the gravitational potential energy of the ball: [1] [3] Next we can look at position [3], at the high point. It’s all gravitational potential energy there, since the ball is not moving (K=0) and the spring is at equilibrium (U elastic = 0). Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

2) A spring-loaded toy gun is used to shoot a ball of mass m=1.50kg straight up in the air, as shown in the figure. The spring has a spring constant of k=667N/m. If the spring is compressed a distance of 25.0cm from its equilibrium position y=0 and then released, find the ball’s maximum height h max (measured from the equilibrium position of the spring.) There is no air resistance, and the ball never touches the inside of the gun. Assume that all movement occurs in a straight line up and down along the y-axis. We will use conservation of energy for this one. Notice that the picture already has y=0 defined for us at the top of the tube (this is also the equilibrium position of the spring). At the beginning (I will call this position [1]) the energy of the system is all potential – the ball is not moving so K=0. We need to account for the compression of the spring, as well as the gravitational potential energy of the ball: [1] [3] Next we can look at position [3], at the high point. It’s all gravitational potential energy there, since the ball is not moving (K=0) and the spring is at equilibrium (U elastic = 0). Conservation of energy says that the total energy should be the same at both points, so E 1 = E 3 = 17.17J Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

3) A block of mass m 1 = 2.40kg is connected to a second block of mass m 2 =1.80kg, as shown on the board. When the blocks are released from rest, they move through a distance d=0.500m, at which point m 2 hits the floor. Give that the coefficient of kinetic friction between m 1 and the horizontal surface is μ k =0.450, find the speed of the blocks just before m 2 lands. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

3) A block of mass m 1 = 2.40kg is connected to a second block of mass m 2 =1.80kg, as shown on the board. When the blocks are released from rest, they move through a distance d=0.500m, at which point m 2 hits the floor. Give that the coefficient of kinetic friction between m 1 and the horizontal surface is μ k =0.450, find the speed of the blocks just before m 2 lands. We can use conservation of energy. Initially nothing is moving, so K=0 and we only have gravitational potential energy. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

3) A block of mass m 1 = 2.40kg is connected to a second block of mass m 2 =1.80kg, as shown on the board. When the blocks are released from rest, they move through a distance d=0.500m, at which point m 2 hits the floor. Give that the coefficient of kinetic friction between m 1 and the horizontal surface is μ k =0.450, find the speed of the blocks just before m 2 lands. We can use conservation of energy. Initially nothing is moving, so K=0 and we only have gravitational potential energy. Just before the block lands, both blocks are moving so we will have kinetic energy, as well as potential energy for block 1 only (block 2 is now at y=0). Also note that both blocks move at the same speed. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

3) A block of mass m 1 = 2.40kg is connected to a second block of mass m 2 =1.80kg, as shown on the board. When the blocks are released from rest, they move through a distance d=0.500m, at which point m 2 hits the floor. Give that the coefficient of kinetic friction between m 1 and the horizontal surface is μ k =0.450, find the speed of the blocks just before m 2 lands. We can use conservation of energy. Initially nothing is moving, so K=0 and we only have gravitational potential energy. Just before the block lands, both blocks are moving so we will have kinetic energy, as well as potential energy for block 1 only (block 2 is now at y=0). Also note that both blocks move at the same speed. In the absence of friction, we would just set these energies equal. We must account for friction by finding the work done by friction, and adding it to the initial energy. Note: this work will be negative, so we will end up with less energy than we started with, as expected (kinetic friction will always take energy away from the system). Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

3) A block of mass m 1 = 2.40kg is connected to a second block of mass m 2 =1.80kg, as shown on the board. When the blocks are released from rest, they move through a distance d=0.500m, at which point m 2 hits the floor. Give that the coefficient of kinetic friction between m 1 and the horizontal surface is μ k =0.450, find the speed of the blocks just before m 2 lands. We can use conservation of energy. Initially nothing is moving, so K=0 and we only have gravitational potential energy. Just before the block lands, both blocks are moving so we will have kinetic energy, as well as potential energy for block 1 only (block 2 is now at y=0). Also note that both blocks move at the same speed. In the absence of friction, we would just set these energies equal. We must account for friction by finding the work done by friction, and adding it to the initial energy. Note: this work will be negative, so we will end up with less energy than we started with, as expected (kinetic friction will always take energy away from the system). The friction work is negative because the force always opposes the motion (that is why the angle is 180 degrees). Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

3) A block of mass m 1 = 2.40kg is connected to a second block of mass m 2 =1.80kg, as shown on the board. When the blocks are released from rest, they move through a distance d=0.500m, at which point m 2 hits the floor. Give that the coefficient of kinetic friction between m 1 and the horizontal surface is μ k =0.450, find the speed of the blocks just before m 2 lands. We can use conservation of energy. Initially nothing is moving, so K=0 and we only have gravitational potential energy. Just before the block lands, both blocks are moving so we will have kinetic energy, as well as potential energy for block 1 only (block 2 is now at y=0). Also note that both blocks move at the same speed. In the absence of friction, we would just set these energies equal. We must account for friction by finding the work done by friction, and adding it to the initial energy. Note: this work will be negative, so we will end up with less energy than we started with, as expected (kinetic friction will always take energy away from the system). The friction work is negative because the force always opposes the motion (that is why the angle is 180 degrees). Finally we just set our final energy equal to the initial energy, plus the (negative) friction work: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

4) A 95.0 kg diver steps off a diving board and drops into the water 3.00m below. At some depth d below the water’s surface, the diver comes to rest. If the non-conservative work done on the diver is W nc = -5120J, what is the depth,d? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

4) A 95.0 kg diver steps off a diving board and drops into the water 3.00m below. At some depth d below the water’s surface, the diver comes to rest. If the non-conservative work done on the diver is W nc = -5120J, what is the depth,d? We will use conservation of energy. First we must define our coordinate system. Using y=0 at the lowest point achieved by the diver, we have the following expressions for the initial and final energy: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

4) A 95.0 kg diver steps off a diving board and drops into the water 3.00m below. At some depth d below the water’s surface, the diver comes to rest. If the non-conservative work done on the diver is W nc = -5120J, what is the depth,d? We will use conservation of energy. First we must define our coordinate system. Using y=0 at the lowest point achieved by the diver, we have the following expressions for the initial and final energy: Notice that the kinetic energy is zero in both places because the speed is zero. The final energy will be the non-conservative work plus the initial energy: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

4) A 95.0 kg diver steps off a diving board and drops into the water 3.00m below. At some depth d below the water’s surface, the diver comes to rest. If the non-conservative work done on the diver is W nc = -5120J, what is the depth,d? We will use conservation of energy. First we must define our coordinate system. Using y=0 at the lowest point achieved by the diver, we have the following expressions for the initial and final energy: Notice that the kinetic energy is zero in both places because the speed is zero. The final energy will be the non-conservative work plus the initial energy: We can solve this for (d+h), then subtract out the given value for h. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Physics 6A Work and Energy examples Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.

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Physics 6A Work and Energy examples Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.

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