Transcript Physics 6B
Physics 6A Work and Energy examples Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Work and Energy Energy comes in many forms. We will most often encounter two kinds of energy: Kinetic Energy – Energy of Motion. The formula is Any moving object has kinetic energy. KE = ½ mv2 Potential Energy – Stored Energy. There will be several types of potential energy: * Gravitational – Energy stored by lifting an object above the earth. more robust formula later, but for now: Ugrav = mgh We will have a * Elastic – Energy stored by stretching or compressing a spring. The formula is Uelastic = ½ kx2 * Electric – Energy stored by charges in an electric field. We will see this next quarter. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Work and Energy 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: KEi + Ui + WNC = KEf + Uf 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 μk=0.2 and μs=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 μk=0.2 and μs=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. Here is the free-body diagram. Since the box is moving horizontally, the only forces that do work are friction and the 400N push. Normal force friction 400 N Calculate the force of kinetic friction: 𝑓𝑘 = 𝜇𝑘 𝑚𝑔 = 0.2 100𝑘𝑔 9.8𝑠𝑚2 10m = 196 𝑁 weight Work done by each force: 𝑊𝑝𝑢𝑠ℎ = 400𝑁 10𝑚 = 4000 𝐽 𝑊𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 = − 196𝑁 10𝑚 = −1960 𝐽 Total work done on box: 𝑊𝑡𝑜𝑡𝑎𝑙 = 2040 𝐽 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 μk=0.2 and μs=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 μk=0.2 and μs=0.4, find the total work done on the box. Does the box move? 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. Normal force friction 30° 400 N weight 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 μk=0.2 and μs=0.4, find the total work done on the box. Does the box move? Normal force First we will need to break the 400N push into components. 𝑥: 400𝑁 𝑐𝑜𝑠30° = 346𝑁 𝑦: 400𝑁 𝑠𝑖𝑛30° = 200𝑁 To get the max friction we need the normal force, which will be greater because of the downward push: 346 N friction 30° 200 N 400 N weight 𝐹𝑛𝑜𝑟𝑚𝑎𝑙 = 200𝑁 + 100𝑘𝑔 9.8𝑠𝑚2 = 1180 𝑁 𝑓𝑠,𝑚𝑎𝑥 = 𝜇𝑠 𝐹𝑛𝑜𝑟𝑚𝑎𝑙 = 0.4 1180 = 472 𝑁 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 holdPrepared the box. 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 freefalls 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 freefalls to the ground instead. We can do this one with conservation of energy. Here is the basic format: KEi + Ui + WNC = KEf + Uf 10m 30° 0 + mgh + 0 = ½ mv2 + 0 Solving for v: 𝑣= 2𝑔ℎ = 2(9.8𝑠𝑚2)(10𝑚) = 14𝑚 𝑠 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 μk=0.2 and μs=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 μk=0.2 and μs=0.3. Find the final speed of the box when it reaches the bottom of the ramp. We can do this one with conservation of energy. Here is the basic format: KEi + Ui + WNC = KEf + Uf 0 + mgh + Wfriction = ½ mv2 + 0 10m 30° 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 μk=0.2 and μs=0.3. Find the final speed of the box when it reaches the bottom of the ramp. 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. mgsin30 mgcos30 Let’s calculate the max static friction and the downhill gravity force to see which is bigger. mg 30° 𝑚𝑔𝑠𝑖𝑛30° = 100𝑘𝑔 9.8𝑠𝑚2 0.5 = 490 𝑁 𝑓𝑠,𝑚𝑎𝑥 = 𝜇𝑠 𝑚𝑔𝑐𝑜𝑠30° = 0.3 100𝑘𝑔 9.8𝑠𝑚2 0.866 = 255 𝑁 Looks like plenty of downhill force to overcome friction. The actual friction will be kinetic: 𝑓𝑘 = 𝜇𝑘 𝑚𝑔𝑐𝑜𝑠30° = 0.2 100𝑘𝑔 9.8𝑠𝑚2 0.866 = 170 𝑁 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 μk=0.2 and μs=0.3. Find the final speed of the box when it reaches the bottom of the ramp. Work done by kinetic friction: 𝑊𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 = −𝑓𝑘 𝑑 = − 170𝑁 20𝑚 = −3400 𝐽 d 10m Now we can fill in the conservation of energy formula. 0 + mgh + Wfriction = ½ mv2 + 0 100𝑘𝑔 9.8𝑠𝑚2 10𝑚 − 3400𝐽 = 12(100𝑘𝑔)(𝑣 2 ) 𝑣 = 11.3 𝑚 𝑠 30° 10𝑚 𝑠𝑖𝑛30° = 𝑑 10𝑚 𝑑= = 20 𝑚 𝑠𝑖𝑛30° 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. 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. F=11N 29° 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. Initially the sled is moving at 0.5 m/s, so its kinetic energy is: 2 K 1 mv2 1 6.4kg 0.5 m 0.8J 2 2 s F=11N 29° 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. Initially the sled is moving at 0.5 m/s, so its kinetic energy is: 2 K 1 mv2 1 6.4kg 0.5 m 0.8J 2 2 s 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. F=11N 29° 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. Initially the sled is moving at 0.5 m/s, so its kinetic energy is: 2 K 1 mv2 1 6.4kg 0.5 m 0.8J 2 2 s 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. F=11N 29° The force is not aligned with the motion, so we need to use the x-component to get the work. W F cos d W 11N cos29 2m 19.24J 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. Initially the sled is moving at 0.5 m/s, so its kinetic energy is: 2 K 1 mv2 1 6.4kg 0.5 m 0.8J 2 2 s 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. F=11N 29° The force is not aligned with the motion, so we need to use the x-component to get the work. W F cos d W 11N cos29 2m 19.24J 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 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. Initially the sled is moving at 0.5 m/s, so its kinetic energy is: 2 K 1 mv2 1 6.4kg 0.5 m 0.8J 2 2 s 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. F=11N 29° The force is not aligned with the motion, so we need to use the x-component to get the work. W F cos d W 11N cos29 2m 19.24J Now the total KE is 20.04J. Use this to solve for the new speed: K 1 mv2 20.04J 2 v2 2(20.04J) v 2.5 m s 6.4kg 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 hmax(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 hmax(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. [3] 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] 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 hmax(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. [3] 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] 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 hmax(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. [3] 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] E1 1 ky12 mgy1 2 E1 1 667 N 0.25m2 1.5kg 9.8 m2 0.25m 2 m s E1 17.17J 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 hmax(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. [3] 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] E1 1 ky12 mgy1 2 E1 1 667 N 0.25m2 1.5kg 9.8 m2 0.25m 2 m s E1 17.17J 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 (Uelastic = 0). E3 mghmax 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 hmax(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. [3] 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] E1 1 ky12 mgy1 2 E1 1 667 N 0.25m2 1.5kg 9.8 m2 0.25m 2 m s E1 17.17J 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 (Uelastic = 0). E3 mghmax Conservation of energy says that the total energy should be the same at both points, so E1 = E3 = 17.17J 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 hmax(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. [3] 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] E1 1 ky12 mgy1 2 E1 1 667 N 0.25m2 1.5kg 9.8 m2 0.25m 2 m s E1 17.17J 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 (Uelastic = 0). E3 mghmax Conservation of energy says that the total energy should be the same at both points, so E1 = E3 = 17.17J mghmax 17.17J hmax 17.17J 1.17m m 1.5kg 9.8 2 s Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 3) A block of mass m1= 2.40kg is connected to a second block of mass m2=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 m2 hits the floor. Give that the coefficient of kinetic friction between m1 and the horizontal surface is μk=0.450, find the speed of the blocks just before m2 lands. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 3) A block of mass m1= 2.40kg is connected to a second block of mass m2=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 m2 hits the floor. Give that the coefficient of kinetic friction between m1 and the horizontal surface is μk=0.450, find the speed of the blocks just before m2 lands. We can use conservation of energy. Initially nothing is moving, so K=0 and we only have gravitational potential energy. Ei m1gh m2gd Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 3) A block of mass m1= 2.40kg is connected to a second block of mass m2=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 m2 hits the floor. Give that the coefficient of kinetic friction between m1 and the horizontal surface is μk=0.450, find the speed of the blocks just before m2 lands. We can use conservation of energy. Initially nothing is moving, so K=0 and we only have gravitational potential energy. Ei m1gh m2gd 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. Ef m1gh 1 m1 m2 v2 2 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 3) A block of mass m1= 2.40kg is connected to a second block of mass m2=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 m2 hits the floor. Give that the coefficient of kinetic friction between m1 and the horizontal surface is μk=0.450, find the speed of the blocks just before m2 lands. We can use conservation of energy. Initially nothing is moving, so K=0 and we only have gravitational potential energy. Ei m1gh m2gd 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. Ef m1gh 1 m1 m2 v2 2 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 m1= 2.40kg is connected to a second block of mass m2=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 m2 hits the floor. Give that the coefficient of kinetic friction between m1 and the horizontal surface is μk=0.450, find the speed of the blocks just before m2 lands. We can use conservation of energy. Initially nothing is moving, so K=0 and we only have gravitational potential energy. Ei m1gh m2gd 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. Ef m1gh 1 m1 m2 v2 2 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). Wfric Ffric d cos180 Wfric k m1g d 1 Wfric k m1g d 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 m1= 2.40kg is connected to a second block of mass m2=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 m2 hits the floor. Give that the coefficient of kinetic friction between m1 and the horizontal surface is μk=0.450, find the speed of the blocks just before m2 lands. We can use conservation of energy. Initially nothing is moving, so K=0 and we only have gravitational potential energy. Ei m1gh m2gd 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. Ef m1gh 1 m1 m2 v2 2 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). Wfric Ffric d cos180 Wfric k m1g d 1 Wfric k m1g d 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: Ef Ei Wfric 1 m m v 2 m gd m g d 2 2 k 1 2 1 v2 m2gd k m1g d v 1.3 m 1 m m s 1 2 2 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 Wnc= -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 Wnc= -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: Ei mg(d h) Ef mg(0) 0 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 Wnc= -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: Ei mg(d h) Ef mg(0) 0 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: Ef Ei Wnc 0 mg(d h) 5120J 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 Wnc= -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: Ei mg(d h) Ef mg(0) 0 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: Ef Ei Wnc 0 mg(d h) 5120J We can solve this for (d+h), then subtract out the given value for h. 0 mg(d h) 5120J 5120J ( d h) 5.5m 95kg 9.8 m2 s d 5.5m 3.0m 2.5m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB