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Energy Problems Multiple Choice A force F of strength 20N acts on an object of mass 3kg as it moves a distance of 4m. If F is perpendicular to the 4m displacement, the work done is equal to: a) b) c) d) e) 0J 60J 80J 600J 2400J Since the Force is perpendicular to the displacement, there is no work done by this force. Multiple Choice Under the influence of a force, an object of mass 4kg accelerates from 3 m/s to 6 m/s in 8s. How much work was done on the object during this time? a) b) c) d) e) 27J 54J 72J 96J Can not be determined from the information given This is a job for the work energy theorem. W K 1 m v 2f vi2 2 m 2 m 2 1 4kg 6 3 s s 2 54 J Multiple Choice A box of mass m slides down a frictionless inclined plane of length L and vertical height h. What is the change in gravitational potential energy? a) b) c) d) e) -mgL -mgh -mgL/h -mgh/L -mghL The length of the ramp is irrelevant, it is only the change in height Multiple Choice An object of mass m is travelling at constant speed v in a circular path of radius r. how much work is done by the centripetal force during one-half of a revolution? a) b) c) d) e) πmv2 J 2πmv2 J 0J πmv2r J 2πmv2/r J Since centripetal force always points along a radius towards the centre of the circle, and the displacement is always along the path of the circle, no work is done. Multiple Choice While a person lifts a book of mass 2kg from the floor to a tabletop, 1.5m above the floor, how much work does the gravitational force do on the book? a) b) c) d) e) -30J -15J 0J 15J 30J The gravitational force points downward, while the displacement is upward W mgh m 2kg 9.8 2 1.5m s 30 J Multiple Choice A block of mass 3.5kg slides down a frictionless inclined plane of length 6m that makes an angle of 30o with the horizontal. If the block is released from rest at the top of the incline, what is the speed at the bottom? a) b) c) d) e) 4.9 m/s 5.2 m/s 6.4 m/s 7.7 m/s 9.2 m/s The work done by gravity is equal to the change in Kinetic Energy. W K 1 mgh m v 2f vi2 2 1 gh v 2f 2 v 2 gh m 2 9.8 2 sin 30 6m s m 7.7 s Multiple Choice A block of mass 3.5kg slides down an inclined plane of length 6m that makes an angle of 60o with the horizontal. The coefficient of kinetic friction between the block and the incline is 0.3. If the block is released from rest at the top of the incline, what is the speed at the bottom? a) b) c) d) e) 4.9 m/s 5.2 m/s 6.4 m/s 7.7 m/s 9.2 m/s Ki U i W f K f U f 1 2 mv f 0 2 1 mg L sin mg cos L v 2f 2 0 mgh Ff L v f 2 gL sin cos This is a conservation of Mechanical Energy, including the negative work done by friction. m 2 9.8 2 6m sin 60 0.3cos 60 s m 9.2 s Multiple Choice An astronaut drops a rock from the top of a crater on the Moon. When the rock is halfway down to the bottom of the crater, its speed is what fraction of its final impact speed? a) b) c) d) e) 1/4 2 1/4 1/2 2 1/2 1/ 2 Total energy is conserved. Since the rock has half of its potential energy, its kinetic energy at the halfway point is half of its kinetic energy at impact. K v2 v 1 2 Multiple Choice A force of 200N is required to keep an object at a constant speed of 2 m/s across a rough floor. How much power is being expended to maintain this motion? a) b) c) d) e) 50W 100W 200W 400W Cannot be determined with given information Use the power equation P Fv m 200 N 2 s 400W Understanding Compare the work done on an object of mass 2.0 kg a) In lifting an object 10.0m b) Pushing it up a ramp inclined at 300 to the same final height pushing lifting 10.0m 300 Understanding Compare the work done on an object of mass 2.0 kg a) In lifting an object 10.0m lifting 10.0m 300 W F d mgh m 2.0kg 9.8 2 10.0m s 196J 200J Understanding Compare the work done on an object of mass 2.0 kg a) In lifting an object 10.0m b) Pushing it up a ramp inclined at 300 to the same final height pushing 10.0m 300 The distance travelled up the ramp d 10.0m 20m sin 30 Applied Force Fapplied mg sin Work W Fapplied d W mg sin d m W 2.0kg 9.8 2 sin 30 20m s W 200J Question A 4.0 kg mass is moving to the right at 2.0 m/s. It collides with a 10.0 kg mass sitting still. Given that the collision is perfectly elastic, determine the final velocities of each of the masses. Solutions Question A 4.0 kg mass is moving to the right at 2.0 m/s. It collides with a 10.0 kg mass sitting still. Given that the collision is perfectly elastic, determine the final velocities of each of the masses. Conservation of Momentum pi p f m1v1i m2v2i m1v1 f m2v2 f 4.0kg 2.0 m 0 4.0kg v1 f 10kg v2 f s kg m 8.0 4.0kg v1 f 10kg v2 f s Conservation of Energy i K2f Oh No, we K have 1 unknowns. 1 1 1 2 2If only 2 2 mthere v m v m v m v 1 1i 2 2 i 1 1 f 2 2f was 2 2 another 2 2 equation. 2 m 4.0kg 2.0 0 4.0kg v 21 f 10kg v 2 2 f s 16.0 J 4.0kg v 21 f 10kg v 2 2 f Question A 4.0 kg mass is moving to the right at 2.0 m/s. It collides with a 10.0 kg mass sitting still. Given that the collision is perfectly elastic, determine the final velocities of each of the masses. 8.0 kg m 4.0kg v1 f 10kg v2 f s kg m 8.0 4.0kg v1 f 10kg v2 f s kg m 8.0 10kg v2 f s v1 f 4.0kg 16.0J 4.0kg v21 f 10kg v22 f kg m 8.0 10kg v2 f s 16.0 J 4.0kg 4.0kg 0 7v22 f 8v2 f 0 v2 f 7v2 f 8 m v2 f 1.1 s 2 2 10kg v 2 f kg m m 10kg 1.14 s s v1 f 4.0kg 0.82 8.0 Example 0 A box slides down an inclined plane (incline angle = 400). The mass of the block, m, is 35 kg, the coefficient of kinetic friction between the box and the ramp, uk, is 0.3, and the length of the ramp, d, is 8m. a) How much work is done by gravity? b) How much work is done by the normal force? c) How much work is done by friction? d) What is the total work done? Solution A box slides down an inclined plane (incline angle = 400). The mass of the block, m, is 35 kg, the coefficient of kinetic friction between the box and the ramp, uk, is 0.3, and the length of the ramp, d, is 8m. a) How much work is done by gravity? Recall Work = force x distance with the force being parallel to the distance, that is, the component down the ramp Fg Wgravity F d mg sin d m 35kg 9.8 2 sin 40 8m s 1760 J Solution A box slides down an inclined plane (incline angle = 400). The mass of the block, m, is 35 kg, the coefficient of kinetic friction between the box and the ramp, uk, is 0.3, and the length of the ramp, d, is 8m. b) How much work is done by the normal force? Since the normal force is perpendicular to the displacement, NO WORK IS DONE. Solution A box slides down an inclined plane (incline angle = 400). The mass of the block, m, is 35 kg, the coefficient of kinetic friction between the box and the ramp, uk, is 0.3, and the length of the ramp, d, is 8m. c) How much work is done by friction? Recall Work = force x distance. W friction Ff d uk mg cos d m 0.3 35kg 9.8 2 cos 40 8m s 630 J Solution A box slides down an inclined plane (incline angle = 400). The mass of the block, m, is 35 kg, the coefficient of kinetic friction between the box and the ramp, uk, is 0.3, and the length of the ramp, d, is 8m. d) What is the total work done? Total Work = sum of all works Wtotal Wgravity WNormal W friction 1760 J 0 J 630 J 1130 J Example 2 A 2.00kg block is pushed against a spring with negligible mass and a force constant k=400 N/m, compressing it 0.220 m. when the block is released, it moves along a frictionless horizontal surface and then up a frictionless incline with a slope of 37.00. a) What is the speed of the block as it slides along the horizontal surface after having left the spring? b) How far does the block travel up the incline before starting to slide back down? Solution to Example 2 A 2.00kg block is pushed against a spring with negligible mass and a force constant k=400 N/m, compressing it 0.220 m. when the block is released, it moves along a frictionless horizontal surface and then up a frictionless incline with a slope of 37.00. a) What is the speed of the block as it slides along the horizontal surface after having left the spring? b) How far does the block travel up the incline before starting to slide back down? Let’s find the Spring Potential Energy first. 1 2 kx 2 1 N 2 400 0.220m 2 m 9.68 J US When released the Spring Potential Energy is transferred into kinetic energy of the Block, where we can solve for v 1 2 mv 2 1 9.68 J 2.00kg v 2 2 EK v 2 9.68 J 2.00kg 3.11 m s Solution to Example 1 A 2.00kg block is pushed against a spring with negligible mass and a force constant k=400 N/m, compressing it 0.220 m. when the block is released, it moves along a frictionless horizontal surface and then up a frictionless incline with a slope of 37.00. a) What is the speed of the block as it slides along the horizontal surface after having left the spring? b) How far does the block travel up the incline before starting to slide back down? We can use conservation of Spring Potential and Gravitational Potential Energy UG U s 1 2 kx 2 mgh 9.68 J mgh 9.68 J m 2.00kg 9.8 2 s 0.494m h Using Trig, we will translate the vertical distance to a ramp distance 0.494m d 0.494m d sin 37.0 sin 37.0 0.821m Question 2 The planet Saturn is moving in the negative x-direction at its orbital speed (with respect to the Sun) of 9.6 km/s. The mass of Saturn is 5.69x1026 kg. A 2150 kg spacecraft approaches Saturn, moving initially in the +x-direction at 10.4 km/s. The gravitational attraction of Saturn (a conservative force) causes the spacecraft to swing around it and head of into the opposite direction. Determine the speed of the spacecraft after it is far enough away to be free of Saturn’s gravitational pull. Solution to Question 2 The planet Saturn is moving in the negative x-direction at its orbital speed (with respect to the Sun) of 9.6 km/s. The mass of Saturn is 5.69x1026 kg. A 2150 kg spacecraft approaches Saturn, moving initially in the +x-direction at 10.4 km/s. The gravitational attraction of Saturn (a conservative force) causes the spacecraft to swing around it and head of into the opposite direction. Determine the speed of the spacecraft after it is far enough away to be free of Saturn’s gravitational pull. Here the “collision” is not an impact but rather a gravitational interaction. So we can treat it as a one-dimensional elastic collision. Therefore the relative velocities before and after the collision have the same magnitude but opposite sign. v pi vsi v pf vsf vsf v pi vsi v pf km km km 10.4 9.6 s s s km 29.6 s 9.6 The reason the spacecraft acquires so much additional speed is that Saturn is moving in its orbit. It does slow down, but its mass is so much greater than the spacecraft, it is not noticeable. Question 3 You and your bicycle have a combined mass of 80.0kg. When you reach the base of an bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction: a) How much work have you done with the force you apply to the pedals? Solution to Question 3 You and your bicycle have a combined mass of 80.0kg. When you reach the base of an bridge, you are traveling along the road at 5.00 m/s. at the top of the bridge, you have climbed a vertical distance of 5.20m and have slowed to 1.50 m/s. Ignoring work done by any friction: a) How much work have you done with the force you apply to the pedals? Conservation of energy states that initial kinetic energy equals final kinetic energy plus work done by gravity less work done by you W ET K f U f Ki U i Wp K f Ki U f U i 1 2 1 2 mv f mvi mgh 0 2 2 2 2 1 m m m 80.0kg 1.50 5.00 80kg 9.8 2 5.2m 2 s s s 3166.8 J 3.17 103 J Question 4 The figure below shows a ballistic pendulum, the system for measuring the speed of a bullet. A bullet of m=1.0 g is fired into a block of wood with mass of M=3.0kg , suspended like a pendulum, and makes a completely inelastic collision with it. After the impact of the bullet, the block swings up to a maximum height 2.00 cm. Determine the initial velocity of the bullet? Stage 1 (conservation of Momentum) pi p f mv1i Mv2i mV MV 0.001kg v1 0 3.001.kg V v1 3001V Stage 2 (conservation of Energy) Ki U f 1 m M V 2 m M gh 2 m V 2 2 9.8 2 0.02m s m V 0.626 s Therefore: m v1 3001 0.626 s m 1879 s Question 5 Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quarter-circle with radius R=3.00m. a) Determine his speed at the bottom of the ramp b) Determine the normal force that acts on him at the bottom of the curve. c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00 m/s. what work was done by the friction force acting on him? Solution to Question 5 Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m. a) Determine his speed at the bottom of the ramp b) Determine the normal force that acts on him at the bottom of the curve. c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00 m/s. what work was done by the friction force acting on him? From Conservation of Energy K1 U1 K 2 U 2 1 2 mv2 0 2 v2 2 gR 0 mgR m 2 9.8 2 3.00m s m 7.67 s Solution to Question 5 Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m. a) Determine his speed at the bottom of the ramp b) Determine the normal force that acts on him at the bottom of the curve. c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00 m/s. what work was done by the friction force acting on him? The free body diagram of the normal is through-out his journey is: F y mac v22 FN FG m R 2 gR FN mg m R mg 2mg 3mg v2 2 gR Solution to Question 5 Cousin Vinney skateboards down a playground ramp. He (25.0 kg) moves through a quartercircle with radius R=3.00m. a) Determine his speed at the bottom of the ramp b) Determine the normal force that acts on him at the bottom of the curve. c) Suppose the ramp is not frictionless and that Vinney’s speed at the bottom is only 6.00 m/s. what work was done by the friction force acting on him? The normal force does no work, but the friction force does do work. Therefore the non-gravitational work done on Vinney is just the work done by friction. The free body diagram of the normal is through-out his journey is: K1 U1 WF K 2 U 2 WF K 2 U 2 K1 U1 1 2 mv2 0 0 mgh 2 2 1 m m 25.0kg 6.00 25.0kg 9.80 2 3.00m 2 s s 285 J Question 6 A 2000 kg elevator with broken cables is falling at 25 m/s when it first contacts a cushioning spring at the bottom of the shaft. The spring is supposed to stop the elevator, compressing 3.00 m as it does. During the motion a safety clamp applies a constant 17,000 N frictional force to the elevator. a) Determine the force constant of the spring. Solution to Question 6 A 2000 kg elevator with broken cables is falling at 25 m/s when it first contacts a cushioning spring at the bottom of the shaft. The spring is supposed to stop the elevator, compressing 3.00 m as it does. During the motion a safety clamp applies a constant 17,000 N frictional force to the elevator. a) Determine the force constant of the spring. The elevator’s initial kinetic energy is: 2 1 1 m K1 mv12 2000kg 25 625, 000 J 2 2 s If we use Point 1 to be the origin, then we have Us=0 and Ugrav=0, and so U1=0 At Point 2, there is both gravitational and elastic potential energy, but no kinetic energy 1 U 2 mgy2 ky22 2 Putting the conservation of energy to work: K1+U1+Wother=K2+U2 Solution to Question 6 A 2000 kg elevator with broken cables is falling at 25 m/s when it first contacts a cushioning spring at the bottom of the shaft. The spring is supposed to stop the elevator, compressing 3.00 m as it does. During the motion a safety clamp applies a constant 17,000 N frictional force to the elevator. a) Determine the force constant of the spring. K1 625,000 J If we use Point 1 to be the origin, then we have Us=0 and Ugrav=0, and so U1=0 1 U 2 mgy2 ky22 2 K1 U1 Wother K 2 U 2 1 625, 000 J 0 J Wother 0 J mgy2 ky22 2 m 2 625, 000 J 17, 000 N 3.00m 2000kg 9.80 2 3.00m s k 2 3.00m 1.41105 N m Example 7 An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine the distance the object travels before coming to a halt. 1.0 kg Solution 7 An object of mass 1.0 kg travelling at 5.0 m/s enters a region of ice where the coefficient of kinetic friction is 0.10. Use the Work Energy Theorem to determine the distance the object travels before coming to a halt. FN Forces We can see that the objects weight is balanced by the normal force exerted by the ice. Therefore the only work done is due to the friction acting on the object. Let’s determine the friction force. F f uk FN uk mg m 0.10 1.0kg 9.8 2 s 0.98N Now apply the work Energy Theorem and solve for d Ff W KE 1 1 Ff d mv 2f mvi2 2 2 mg 2 1 m 1 m 0.98 N d 1.0kg 0 1.0kg 5.0 2 s s 2 12.5 J d 0.98 N 13m 2 Example 8 A box of mass m is released from rest at point A, the top of a long frictionless slide. Point A is at height H above the level points B and C. Although the slide is frictionless, the horizontal surface from point B to C is not. The coefficient of kinetic friction between the box and this surface is uk, and the horizontal distance between points B and C is x. A a) b) c) d) e) Find the speed of the box when its height above the ground is 1/2H Find the speed of the box when it reaches B. Determine the value of uk, so that it comes to a rest at C Determine the value of uk if C was at a height of h above the ground. If the slide was not frictionless, determine the work done by friction as the box moved from A to B if the speed at B was ½ of the speed calculated in b) H uk B x C Example 8 A box of mass m is released from rest at point A, the top of a long frictionless slide. Point A is at height H above the level points B and C. Although the slide is frictionless, the horizontal surface from point B to C is not. The coefficient of kinetic friction between the box and this surface is uk, and the horizontal distance between points B and C is x. a) Find the speed of the box when its height above the ground is 1/2H Etotal UG K mgH 0 mgH A H U 1 G 2 K 1 Etotal 2 1 1 2 mg H mv mgH 2 2 1 2 1 mv mgH 2 2 v gH uk B x C Example 8 A box of mass m is released from rest at point A, the top of a long frictionless slide. Point A is at height H above the level points B and C. Although the slide is frictionless, the horizontal surface from point B to C is not. The coefficient of kinetic friction between the box and this surface is uk, and the horizontal distance between points B and C is x. b) Find the speed of the box when it reaches B. Etotal UG K mgH 0 mgH A U GB K B Etotal 1 2 0 mv mgH 2 v 2 2 gH H v 2 gH uk B x C Example 8 A box of mass m is released from rest at point A, the top of a long frictionless slide. Point A is at height H above the level points B and C. Although the slide is frictionless, the horizontal surface from point B to C is not. The coefficient of kinetic friction between the box and this surface is uk, and the horizontal distance between points B and C is x. c) Determine the value of uk, so that it comes to a rest at C A H W K 1 2 1 2 W mv f mvi 2 2 1 2 mvi 2 1 2 Fd mvi 2 1 2 mguk x mvi 2 v 2 gH vi2 uk 2 gx H x uk B x C Example 8 A box of mass m is released from rest at point A, the top of a long frictionless slide. Point A is at height H above the level points B and C. Although the slide is frictionless, the horizontal surface from point B to C is not. The coefficient of kinetic friction between the box and this surface is uk, and the horizontal distance between points B and C is x. d) Determine the value of uk if C was at a height of h above the ground. K B U B W f KC U C mgH 0 Ff L 0 mgh A H h uk L cos mgH uk mg cos L mgh H H uk cos L h L B H h x uk x C Example 8 A box of mass m is released from rest at point A, the top of a long frictionless slide. Point A is at height H above the level points B and C. Although the slide is frictionless, the horizontal surface from point B to C is not. The coefficient of kinetic friction between the box and this surface is uk, and the horizontal distance between points B and C is x. e) If the slide was not frictionless, determine the work done by friction as the box moved from A to B if the speed at B was ½ of the speed calculated in b) A H uk U A K A Wf U B KB 1 2 U A K A W f mvB 2 2 1 1 mgH 0 W f m 2 gH 2 2 mgH mgH W f 4 mgH 3 Wf mgH mgH 4 4 B x Example 9 An acrobat swings from the horizontal. When the acrobat was swung an angle of 300, what is his velocity at that point, if the length of the rope is L? Because gravity is a conservative force (the work done by the rope is tangent to the motion of travel), Gravitational Potential Energy is converted to Kinetic Energy. It’s too difficult to use Centripetal Acceleration UI K f U f 1 2 mv mgL sin 30 2 1 1 mgL mv 2 2 2 gL v 2 mgL v gL 30 L h L sin 30 Example 10 A block of mass 3.00 kg sits on a horizontal frictionless surface. It is attached to a spring with a spring constant of 36.0 N / m. The weight is pulled 0.16 m away from the equilibrium point and then released. a) Assuming no damping, what is the speed of the block, i) as it passes through the equilibrium point? ii) when it has moved 5.0 cm from its release point? b) Find the time after release that the block first passed through the position x = 3.0 cm (on the opposite side of its equilibrium point). 0.16m Example 10 Solution A block of mass 3.00 kg sits on a horizontal frictionless surface. It is attached to a spring with a spring constant of 36.0 N / m. The weight is pulled 0.16 m away from the equilibrium point and then released. a) Assuming no damping, what is the speed of the block, i) as it passes through the equilibrium point? We can solve this in two different ways 1) Using Energies 1 2 1 2 kA mv 2 2 0.16m 1 2 1 2 kA mv 2 2 k v A m N m 0.160m 3.00kg 36.0 0.554 m s Example 10 Solution A block of mass 3.00 kg sits on a horizontal frictionless surface. It is attached to a spring with a spring constant of 36.0 N / m. The weight is pulled 0.16 m away from the equilibrium point and then released. a) Assuming no damping, what is the speed of the block, i) as it passes through the equilibrium point? We can solve this in two different ways 2) Using Velocity Function k k v A sin t m m 0.16m ¼ of T T 2 We need only the time when the mass is at the equilibrium point (ie x=0). We can get that by setting the position function = 0 and solving for time or by using one quarter of the period. m k v 0.16 3 36 0.16 1 2 T 4 4 t 1 4 3 36 3 m 0.554 s 36 1 36 sin 3 4 3 3 Example 10 A block of mass 3.00 kg sits on a horizontal frictionless surface. It is attached to a spring with a spring constant of 36.0 N / m. The weight is pulled 0.16 m away from the equilibrium point and then released. a) Assuming no damping, what is the speed of the block, ii) when it has moved 5.0 cm from its release point? We can solve this in two different ways 1) Using Energies 1 2 1 2 1 2 kA mv kx 2 2 2 5cm 1 2 1 2 1 2 kx mv kA 2 2 2 k 2 v A x2 m x is the position that the spring is stretched from the equilibrium at that at that velocity. This distance is 16cm – 5 cm= 11 cm. N m 3.00kg 36.0 0.40 m s 16 cm 0.160m 0.110m 2 2 Example 10 A block of mass 3.00 kg sits on a horizontal frictionless surface. It is attached to a spring with a spring constant of 36.0 N / m. The weight is pulled 0.16 m away from the equilibrium point and then released. a) Assuming no damping, what is the speed of the block, ii) when it has moved 5.0 cm from its release point? We can solve this in two different ways 2) Using Velocity Function 5cm 36 0.11 0.16 cos t k k 3 16 cm v A sin t m 36 m cos t 0.6875 3 v 0.16 36 sin 36 0.2346 3 3 Setting the position functions to 0.11 m and 36 t cos 1 0.6875 solving for time 3 m k x A cos t m 36 t 0.812755 3 t 0.2346 0.40 s Example 10 A block of mass 3.00 kg sits on a horizontal frictionless surface. It is attached to a spring with a spring constant of 36.0 N / m. The weight is pulled 0.16 m away from the equilibrium point and then released. b) Find the time after release that the block first passed through the position x = 3.0 cm (on the opposite side of its equilibrium point). This is a job for the position function. k x A cos t m 3 cm k x A cos t m 36 0.030m 0.160m cos t 3 cos 12t 0.188 12t cos 1 0.188 t 0.508s 16 cm Example 14 The diagram below shows a roller-coaster ride which contains a circular loop of radius r. A car (mass m) begins from rest from Point A and moves down a frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling once around the circle (B to C to D to E back to B) it then travels along a flat portion from B to F (which is not frictionless). a) Find the centripetal acceleration of the car when it is at Point C b) Determine the speed of the car when its position relative to B is specified by the angle θ shown in the diagram. c) What is the minimum cut-off speed vc that the car must have at D to make it around the loop? d) What is the minimum height H necessary to ensure that the car makes it around the loop? e) If H=6r and the coefficient of friction between the car and the flat portion of the track from B to F is 0.5, how far along the flat portion of the track will the car travel before coming to rest at F? A D H E θ B C F Example 14 The diagram below shows a roller-coaster ride which contains a circular loop of radius r. A car (mass m) begins from rest from Point A and moves down a frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling once around the circle (B to C to D to E back to B) it then travels along a flat portion from B to F (which is not frictionless). a) Find the centripetal acceleration of the car when it is at Point C A K A U A KC U c D H E The centripetal acceleration is v2/r θ B 0 mgH C F 1 2 mvC mgr 2 vC2 2 g H r vC2 2 g H r r r 2g H r aC r Example 14 The diagram below shows a roller-coaster ride which contains a circular loop of radius r. A car (mass m) begins from rest from Point A and moves down a frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling once around the circle (B to C to D to E back to B) it then travels along a flat portion from B to F (which is not frictionless). b) Determine the speed of the car when its position relative to B is specified by the A angle θ shown in the diagram. KA U A K U D H E θ B 0 mgH C 1 2 mv mg r r cos 180 2 F 1 2 mv mg r r cos 2 1 mg H r 1 cos mv 2 2 0 mgH The car’s height above the bottom of the track is given by h=r+rcos(1800-θ). v 2 g H r 1 cos Example 14 The diagram below shows a roller-coaster ride which contains a circular loop of radius r. A car (mass m) begins from rest from Point A and moves down a frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling once around the circle (B to C to D to E back to B) it then travels along a flat portion from B to F (which is not frictionless). c) What is the minimum cut-off speed vc that the car must have at D to make it A around the loop? D H E θ B FC FG FN C When the car reaches D, the forces acting on the car are its weight, mg, and the downward Normal Force. These force just match the Centripetal Force with the Normal equalling zero at the cut-off velocity. F v2 m mg 0 r vcut off rmg m gr Example 14 The diagram below shows a roller-coaster ride which contains a circular loop of radius r. A car (mass m) begins from rest from Point A and moves down a frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling once around the circle (B to C to D to E back to B) it then travels along a flat portion from B to F (which is not frictionless). d) What is the minimum height H necessary to ensure that the car makes it around A the loop? K A U A KD U D D H E C B 1 2 0 mgH mv mg 2r 2 F mgH Apply the cut-off speed from c) to the conservation of Mechanical Energy. 1 m gr mg 2r 2 5 mgr 2 H 5 r 2 Example 14 The diagram below shows a roller-coaster ride which contains a circular loop of radius r. A car (mass m) begins from rest from Point A and moves down a frictionless ramp to Point B where it enters a vertical loop (frictionless), travelling once around the circle (B to C to D to E back to B) it then travels along a flat portion from B to F (which is not frictionless). e) If H=6r and the coefficient of friction between the car and the flat portion of the track from B to F is 0.5, how far along the flat portion of the track will the car travel before coming to rest at F? K A U A KB U B A 1 0 mg (6r ) mvB 2 0 2 D H E C B F Ff x 6mgr Calculate the Kinetic Energy at B, then determine how much work friction must do to remove this energy. k mgx 6mgr 6r 6r x 12r k 0.5