Slide 1 / 76. Slide 2 / 76. Slide 3 / 76. Work & Energy Multiple Choice Problems A 1,800 B 5,000 E 300,000. A Fdcos θ - μ mgd B Fdcos θ.
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1 Slide 1 / 76 Work & nergy Multiple hoice Problems 1 driver in a 2000 kg Porsche wishes to pass a slow moving school bus on a 4 lane road. What is the average power in watts required to accelerate the sports car from 30 m/s to 60 m/s in 9 seconds? Slide 2 / 76 1,800 5,000 10, , ,000 2 force F is at an angle θ above the horizontal is used to pull a heavy suitcase of weight mg a distance d along a level floor at constant velocity. The coefficient of friction between the floor and the suitcase is μ. The work done by the force F is: Slide 3 / 76 Fdcos θ - μ mgd Fdcos θ -μ mgd 2Fdsin θ - μ mgd Fdcos θ - 1
2 3 force of 20 N compresses a spring with spring constant 50 N/m. How much energy is stored in the spring? Slide 4 / 76 2 J 5 J 4 J 6 J 8 J 4 stone is dropped from the edge of a cliff. Which of the following graphs best represents the stone's kinetic energy K as a function of time t? Slide 5 / kg ball is attached to a 1.5 m long string and whirled in a horizontal circle at a constant speed of 5 m/s. How much work is done on the ball during one period Slide 6 / 76 9 J 4.5 J zero 2 J 8 J
3 6 student pushes a box across a horizontal surface at a constant speed of 0.6 m/s. The box has a mass of 40 kg, and the coefficient of kinetic friction is 0.5. The power supplied to the box by the person is Slide 7 / W 60 W 150 W 120 W 200 W 7 force F is applied in horizontal to a 10 kg block. The block moves at a constant speed 2 m/s across a horizontal surface. The coefficient of kinetic friction between the block and the surface is 0.5. The work done by the force F in 1.5 minutes is: Slide 8 / J 5000 J 3000 J 2000 J 1000 J 8 ball swings from point 1 to point 3. ssuming the ball is in SHM and point 3 is 2 m above the lowest point 2. nswer the following questions. What happens to the kinetic energy of the ball when it moves from point 1 to point 2? Slide 9 / 76 increases decreases remains the same zero more information is required
4 9 ball swings from point 1 to point 3. ssuming the ball is in SHM and point 3 is 2 m above the lowest point 2. nswer the following questions. What is the velocity of the ball at the lowest point 2? Slide 10 / m/s 3.5 m/s 4.7 m/s 5.1 m/s 6.3 m/s 10 block with a mass of m slides at a constant velocity V 0 on a horizontal frictionless surface. The block collides with a spring and comes to rest when the spring is compressed to the maximum value. If the spring constant is K, what is the maximum compression in the spring? Slide 11 / 76 V 0 (m/k) 1/2 KmV 0 V 0 K/m m V 0 /K V 0 (K/m) 1/2 F (V 0 m/k) 1/ kg block released from rest from the top of an incline plane. There is no friction between the block and the surface. How much work is done by the gravitational force on the block? Slide 12 / J 60 J 50 J 40 J 30 J
5 12 2 kg block released from rest from the top of an incline plane. There is no friction between the block and the surface. What is the speed of the block when it reaches the horizontal surface? Slide 13 / m/s 4.3 m/s 5.8 m/s 7.7 m/s 6.6 m/s 13 crane lifts a 300 kg load at a constant speed to the top of a building 60 m high in 15 s. The average power expended by the crane to overcome gravity is: Slide 14 / 76 10,000 W 12,000 W 15,000 W 30,000 W 60,000 W 14 satellite with a mass m revolves around arth in a circular orbit with a constant radius R. What is the kinetic energy of the satellite if arth s mass is M? Slide 15 / 76 ½ mv mgh ½GMm/R 2 ½ GMm/R 2Mm/R
6 15 n apple of mass m is thrown in horizontal from the edge of a cliff with a height of H. What is the total mechanical energy of the apple with respect to the ground when it is at the edge of the cliff? Slide 16 / /2mv 0 mgh ½ mv mgh 2 mgh - ½ mv 0 2 mgh + ½ mv 0 16 n apple of mass m is thrown in horizontal from the edge of a cliff with a height of H. What is the kinetic energy of the apple just before it hits the ground? Slide 17 / 76 ½ mv mgh ½ mv mgh mgh 2 ½ mv 0 2 mgh - 1/2 mv kg roller coaster car starts from rest at point and moves down the curved track. Ignore any energy loss due to friction. Find the speed of the car at the lowest point. Slide 18 / m/s 20 m/s 30 m/s 40 m/s 50 m/s
7 kg roller coaster car starts from rest at point and moves down the curved track. Ignore any energy loss due to friction. Find the speed of the car when it reaches point. Slide 19 / m/s 20 m/s 30 m/s 40 m/s 50 m/s 19 Two projectiles and are launched from the ground with velocities of 50 m/s at 60 and 50 m/s at 30 with respect to the horizontal. ssuming there is no air resistance involved, which projectile has greater kinetic energy when it reaches the highest point? Slide 20 / 76 projectile projectile they both have the same none zero kinetic energy they both have zero kinetic energy at the highest point more information is required 20 n object with a mass of 2 kg is initially at rest at a position x = 0. non-constant force F is applied to the object over 6 meters. What is the total work done on the object? Slide 21 / J 150 J 170 J 190 J 180 J
8 21 n object with a mass of 2 kg is initially at rest at a position x=0. non-constant force F is applied to the object over 6 meters. What is the velocity of the object at 6 meters? Slide 22 / m/s 25 m/s 300 m/s m/s not enough information 22 metal ball is held stationary at a height h 0 above the floor and then thrown upward. ssuming the collision with the floor is elastic, which graph best shows the relationship between the total energy of the metal ball and its height h with respect to the floor? Slide 23 / toy car travels with speed v o at point x. Point Y is a height H below point x. ssuming there is no frictional losses and no work is done by a motor, what is the speed at point Y? Slide 24 / 76 (2gH+1/2v 2 o ) 1/2 v o -2gH (2gH + v 2 o ) 1/2 2gH+(1/2v 2 o ) 1/2 v o +2gH
9 24 rocker is launched from the surface of a planet with mass M and radius R. What is the minimum velocity the rocket must be given to completely escape from the planet's gravitational field? Slide 25 / 76 (2GM/R) 1/2 (2GM/R) 1/2 (GM/R) 1/2 2GM/R 2GM/16R 2 25 block of mass m is placed on a frictionless inclined plane with an incline angle θ. The block is just in contact with a free end of an unstretched spring with a spring constant k. If the block is released from rest, what is the maximum compression in the spring? Slide 26 / 76 kmg sinθ kmg cosθ 2mg sinθ/k mg/k kmg 26 In a physics lab a student uses three light frictionless PSO lab carts. ach cart is loaded with some blocks, each having the same mass. The same force F is applied to each cart and they move equal distances d. In which on of these three cases is more work done by force F? Slide 27 / 76 cart I cart II cart III the same work is done on each more information is required
10 27 In a physics lab a student uses three light frictionless PSO lab carts. ach cart is loaded with some blocks, each having the same mass. The same force F is applied to each cart and they move equal distances d. Which cart will have more kinetic energy at the end of distance d? Slide 28 / 76 cart I cart II cart III all three will have the same kinetic energy more information is required 28 In a physics lab a student uses three light frictionless PSO lab carts. ach cart is loaded with some blocks, each having the same mass. The same force F is applied to each cart and they move equal distances d. Which cart will move faster at the end of distance d? Slide 29 / 76 cart I cart II cart III all three will move with the same speed more information is required Slide 30 / 76
11 30 ball of mass m is fastened to a string. The ball swings in a vertical circle of radius r with the other end of the string held fixed. Neglecting air resistance, the difference between the string's tension at the bottom of the circle and at the top of the circle is: Slide 31 / 76 mg 2mg 3mg 6mg 9mg Slide 32 / 76 Work & nergy Free Response Problems kg block is pulled from rest by a force of 1000 N at 37 across a horizontal rough surface over a distance of 5.6 m. The coefficient of kinetic friction between the block and the surface is 0.5. Slide 33 / 76 a. raw a free-body diagram and show all the applied forces. b. How much work is done by force F? c. How much work is done by the normal force? d. How much work is done by the gravitational force? e. How much work is done by the friction force? f. What is the net work done on the block? g. What is the change in kinetic energy of the block?
12 1. 50 kg block is pulled from rest by a force of 1000 N at 37 across a horizontal rough surface over a distance of 5.6 m. The coefficient of kinetic friction between the block and the surface is 0.5. Slide 34 / 76 a. raw a free-body diagram and show all the applied forces. F N F f mg kg block is pulled from rest by a force of 1000 N at 37 across a horizontal rough surface over a distance of 5.6 m. The coefficient of kinetic friction between the block and the surface is 0.5. Slide 35 / 76 b. How much work is done by force F? W = FΔx(os θ) W = (1000 N)(5.6 m)(os37) W = 4472 J kg block is pulled from rest by a force of 1000 N at 37 across a horizontal rough surface over a distance of 5.6 m. The coefficient of kinetic friction between the block and the surface is 0.5. Slide 36 / 76 c. How much work is done by the normal force? 0 J
13 1. 50 kg block is pulled from rest by a force of 1000 N at 37 across a horizontal rough surface over a distance of 5.6 m. The coefficient of kinetic friction between the block and the surface is 0.5. Slide 37 / 76 d. How much work is done by the gravitational force? 0 J kg block is pulled from rest by a force of 1000 N at 37 across a horizontal rough surface over a distance of 5.6 m. The coefficient of kinetic friction between the block and the surface is 0.5. Slide 38 / 76 e. How much work is done by the friction force? ΣF = ma F N + FSinθ = mg F N = mg - FSinθ f = μfn f = μ(mg - FSinθ) f = (0.5)[(50kg)(9.8m/s 2 ) N Sin37 ] = N W = FΔx = (214.8 N)(5.6 m) = 1203 J kg block is pulled from rest by a force of 1000 N at 37 across a horizontal rough surface over a distance of 5.6 m. The coefficient of kinetic friction between the block and the surface is 0.5. Slide 39 / 76 f. What is the net work done on the block? W net = F netδx = (Fsinθ-fΔx) W net = [(1000 N)(sin37 ) N)](5.6 m) W net = 2167 J
14 1. 50 kg block is pulled from rest by a force of 1000 N at 37 across a horizontal rough surface over a distance of 5.6 m. The coefficient of kinetic friction between the block and the surface is 0.5. Slide 40 / 76 g. What is the change in kinetic energy of the block? ΔK = W app - W f ΔK = 4,472J - 1,203J = 3,269J 2. boy pushes a 10 kg sled at a constant speed by applying a force of 75 N at 30 with respect to the horizontal. The sled is pushed over a distance of 15 m. a. raw a free-body diagram and show all the applied forces. b. How much work is done by force F? c. How much work is done by the normal force? Slide 41 / 76 d. How much work is done by the gravitational force? e. How much work is done by the friction force? f. What is the coefficient of kinetic friction between the sled and the surface? g. How much work is done by the net force on the sled? 2. boy pushes a 10 kg sled at a constant speed by applying a force of 75 N at 30 with respect to the horizontal. The sled is pushed over a distance of 15 m. a. raw a free-body diagram and show all the applied forces. Slide 42 / 76 F N mg F
15 2. boy pushes a 10 kg sled at a constant speed by applying a force of 75 N at 30 with respect to the horizontal. The sled is pushed over a distance of 15 m. Slide 43 / 76 b. How much work is done by force F? W = FΔx osθ W = (75 N)(15 m)(os 30 ) W = 974 J 2. boy pushes a 10 kg sled at a constant speed by applying a force of 75 N at 30 with respect to the horizontal. The sled is pushed over a distance of 15 m. Slide 44 / 76 c. How much work is done by the normal force? 0 J 2. boy pushes a 10 kg sled at a constant speed by applying a force of 75 N at 30 with respect to the horizontal. The sled is pushed over a distance of 15 m. Slide 45 / 76 d. How much work is done by the gravitational force? 0 J
16 2. boy pushes a 10 kg sled at a constant speed by applying a force of 75 N at 30 with respect to the horizontal. The sled is pushed over a distance of 15 m. Slide 46 / 76 e. How much work is done by the friction force? W f = -974 J 2. boy pushes a 10 kg sled at a constant speed by applying a force of 75 N at 30 with respect to the horizontal. The sled is pushed over a distance of 15 m. Slide 47 / 76 f. What is the coefficient of kinetic friction between the sled and the surface? W = fδx W = μ(mg + FSinθ) μ = W/(mg + FSinθ)Δx μ = 974J/[(10kg)(9.8m/s 2 ) + 75Sin30 ](15m) = boy pushes a 10 kg sled at a constant speed by applying a force of 75 N at 30 with respect to the horizontal. The sled is pushed over a distance of 15 m. Slide 48 / 76 g. How much work is done by the net force on the sled? 0 J
17 3. 5 kg block is released from rest at the top of a quarter-circle type curved frictionless surface. The radius of the curvature is 3.8 m. When the block reaches the bottom of the curvature it then slides on a rough horizontal surface until it comes to rest. The coefficient of kinetic friction on the horizontal surface is Slide 49 / 76 a. What is the kinetic energy of the block at the bottom of the curved surface? b. What is the speed of the block at the bottom of the curved surface? c. Find the stopping distance of the block? d. Find the elapsed time of the block while it is moving on the horizontal part of the track. e. How much work is done by the friction force on the block on the horizontal part of the track? 3. 5 kg block is released from rest at the top of a quarter-circle type curved frictionless surface. The radius of the curvature is 3.8 m. When the block reaches the bottom of the curvature it then slides on a rough horizontal surface until it comes to rest. The coefficient of kinetic friction on the horizontal surface is Slide 50 / 76 a. What is the kinetic energy of the block at the bottom of the curved surface? 0 + W = f GP = K mgh = K K = mgh = (5kg)(9.8m/s 2 )(3.8m) K = 186J 3. 5 kg block is released from rest at the top of a quarter-circle type curved frictionless surface. The radius of the curvature is 3.8 m. When the block reaches the bottom of the curvature it then slides on a rough horizontal surface until it comes to rest. The coefficient of kinetic friction on the horizontal surface is Slide 51 / 76 b. What is the speed of the block at the bottom of the curved surface? K = ½mv 2 v = (2K/m) 1/2 v = (2)(186J)/5kg) 1/2 v = 8.6 m/s 2
18 3. 5 kg block is released from rest at the top of a quarter-circle type curved frictionless surface. The radius of the curvature is 3.8 m. When the block reaches the bottom of the curvature it then slides on a rough horizontal surface until it comes to rest. The coefficient of kinetic friction on the horizontal surface is Slide 52 / 76 c. Find the stopping distance of the block? 0 + W = f K - W = 0 K = W K = fδx Δx = K/f = K/μmg Δx = 186J/(0.02)(5kg)(9.8m/s 2 ) = 190 m 3. 5 kg block is released from rest at the top of a quarter-circle type curved frictionless surface. The radius of the curvature is 3.8 m. When the block reaches the bottom of the curvature it then slides on a rough horizontal surface until it comes to rest. The coefficient of kinetic friction on the horizontal surface is Slide 53 / 76 d. Find the elapsed time of the block while it is moving on the horizontal part of the track. a = F/m = -f/m v = vo + at -vo = at -vo = (- f/m)t t = vom/f t = vom/μmg t = vo/μg = (8.6m/s)/(0.2)(9.8m/s2) = 44s 3. 5 kg block is released from rest at the top of a quarter-circle type curved frictionless surface. The radius of the curvature is 3.8 m. When the block reaches the bottom of the curvature it then slides on a rough horizontal surface until it comes to rest. The coefficient of kinetic friction on the horizontal surface is Slide 54 / 76 e. How much work is done by the friction force on the block on the horizontal part of the track? 186 J
19 4. Spring gun with a spring constant K is placed at the edge of a table which distance above the floor is H and the apparatus is used to shoot marbles with a certain initial speed in horizontal. The spring is initially compressed by a distance X and then released. The mass of each marble is m. a. How much work is done by the spring on the marble? b. What is the speed of the marble at the edge of the table? c. What is the total energy of the marble at the edge of the table with respect to floor level? d. How much time it will take the marble to reach the floor level from the table? e. What is the horizontal range of the marble? Slide 55 / 76 f. What is the kinetic energy of the marble just before it strikes the floor? 4. Spring gun with a spring constant K is placed at the edge of a table which distance above the floor is H and the apparatus is used to shoot marbles with a certain initial speed in horizontal. The spring is initially compressed by a distance X and then released. The mass of each marble is m. a. How much work is done by the spring on the marble? Slide 56 / 76 W = P = ½kx 2 4. Spring gun with a spring constant K is placed at the edge of a table which distance above the floor is H and the apparatus is used to shoot marbles with a certain initial speed in horizontal. The spring is initially compressed by a distance X and then released. The mass of each marble is m. Slide 57 / 76 b. What is the speed of the marble at the edge of the table? ½KX 2 = ½mv 2 v 2 = KX 2 /m v = X(K/m) 1/2
20 4. Spring gun with a spring constant K is placed at the edge of a table which distance above the floor is H and the apparatus is used to shoot marbles with a certain initial speed in horizontal. The spring is initially compressed by a distance X and then released. The mass of each marble is m. Slide 58 / 76 c. What is the total energy of the marble at the edge of the table with respect to floor level? = ½mv 2 + mgh = ½m(KX 2 /m) + mgh = ½KX 2 + mgh 4. Spring gun with a spring constant K is placed at the edge of a table which distance above the floor is H and the apparatus is used to shoot marbles with a certain initial speed in horizontal. The spring is initially compressed by a distance X and then released. The mass of each marble is m. Slide 59 / 76 d. How much time it will take the marble to reach the floor level from the table? H = ½gt 2 t = (2H/g) 1/2 4. Spring gun with a spring constant K is placed at the edge of a table which distance above the floor is H and the apparatus is used to shoot marbles with a certain initial speed in horizontal. The spring is initially compressed by a distance X and then released. The mass of each marble is m. Slide 60 / 76 e. What is the horizontal range of the marble? R = v o t R = x(k/m) 1/2 (2H/g) 1/2 R = (2HKx 2 /mg) 1/2
21 4. Spring gun with a spring constant K is placed at the edge of a table which distance above the floor is H and the apparatus is used to shoot marbles with a certain initial speed in horizontal. The spring is initially compressed by a distance X and then released. The mass of each marble is m. Slide 61 / 76 f. What is the kinetic energy of the marble just before it strikes the floor? K = ½KX 2 + mgh 5. 5 kg object is initially at rest at x 0 = 0. nonconstant force is applied to the object. The applied force as a function of position is shown on the graph. Slide 62 / 76 a. How much work is done on the object during first 12.5 m? b. What is the change is kinetic energy at the end of 12.5 m? c. What is the speed of the object at the end of 12.5 m? d. What is the total work done by the force for the entire trip? e. What is the change in kinetic energy for the entire trip? f. What is the speed of the object at the end of 20 m? 5. 5 kg object is initially at rest at x 0 = 0. nonconstant force is applied to the object. The applied force as a function of position is shown on the graph. Slide 63 / 76 a. How much work is done on the object during first 12.5 m? W = area under the F vs. x graph W =½bh W =½(12.5m)(40N) W = 250J
22 5. 5 kg object is initially at rest at x 0 = 0. nonconstant force is applied to the object. The applied force as a function of position is shown on the graph. Slide 64 / 76 b. What is the change is kinetic energy at the end of 12.5 m? ΔK = W = 250J 5. 5 kg object is initially at rest at x 0 = 0. nonconstant force is applied to the object. The applied force as a function of position is shown on the graph. Slide 65 / 76 c. What is the speed of the object at the end of 12.5 m? K = ½mv 2 2K/m = v 2 v = (2K/m) 1/2 v = (2(250J)/5kg) 1/2 v = 10 m/s 5. 5 kg object is initially at rest at x 0 = 0. nonconstant force is applied to the object. The applied force as a function of position is shown on the graph. Slide 66 / 76 d. What is the total work done by the force for the entire trip? W = area under the F vs. x graph W =½bh + ½bh W =½(12.5m)(40N) + ½(7.5m)(40N) W = 250J + 150J W = 400J
23 5. 5 kg object is initially at rest at x 0 = 0. nonconstant force is applied to the object. The applied force as a function of position is shown on the graph. Slide 67 / 76 e. What is the change in kinetic energy for the entire trip? ΔK = W = 400J 5. 5 kg object is initially at rest at x 0 = 0. nonconstant force is applied to the object. The applied force as a function of position is shown on the graph. Slide 68 / 76 f. What is the speed of the object at the end of 20 m? K = ½mv 2 2K/m = v 2 v = (2K/m) 1/2 v = (2(400J)/5kg) 1/2 v = 12.6 m/s kg roller coaster car starts from rest at point rolls down the track and then goes around a loop and when it leaves the loop flies off the inclined part of the track. ll the dimensions are: H = 80 m, r = 15 m, h = 10 m, ϴ = 30. a. What is the speed of the car at point? Slide 69 / 76 b. What is the speed of the car at point? c. What is the speed of the car at point? d. What is the force applied by the surface on the car at point? e. What is the force applied by the surface on the car at point? f. How far from point will the car land?
24 kg roller coaster car starts from rest at point rolls down the track and then goes around a loop and when it leaves the loop flies off the inclined part of the track. ll the dimensions are: H = 80 m, r = 15 m, h = 10 m, ϴ = 30. a. What is the speed of the car at point? Slide 70 / 76 GP = K mgh = ½mv 2 gh = ½v 2 v = (2gh) 1/2 v = (2(9.8m/s 2 )(80m)) 1/2 v = 39.6 m/s kg roller coaster car starts from rest at point rolls down the track and then goes around a loop and when it leaves the loop flies off the inclined part of the track. ll the dimensions are: H = 80 m, r = 15 m, h = 10 m, ϴ = 30. Slide 71 / 76 b. What is the speed of the car at point? GP = K + GP f mgh = ½mv 2 + mg2r gh = ½v 2 + g2r v 2 = 2g(H - 2r) v = (2g(H - 2r)) 1/2 v = (2(9.8m/s 2 )(80m-30m)) 1/2 v = 31.3 m/s kg roller coaster car starts from rest at point rolls down the track and then goes around a loop and when it leaves the loop flies off the inclined part of the track. ll the dimensions are: H = 80 m, r = 15 m, h = 10 m, ϴ = 30. Slide 72 / 76 c. What is the speed of the car at point? GP = K + GP f mgh = ½mv 2 + mgh gh = ½v 2 + gh v 2 = 2g(H - h) v = (2g(H - h)) 1/2 v = (2(9.8m/s 2 )(80m-10m)) 1/2 v = 37.0 m/s
25 kg roller coaster car starts from rest at point rolls down the track and then goes around a loop and when it leaves the loop flies off the inclined part of the track. ll the dimensions are: H = 80 m, r = 15 m, h = 10 m, ϴ = 30. Slide 73 / 76 d. What is the force applied by the surface on the car at point? F N - mg = mv 2 /r F N = mg + mv 2 /r F N = m(g + v 2 /r) F N = 900kg(9.8m/s 2 + (39.6m/s) 2 /15m) F N = 102,910N kg roller coaster car starts from rest at point rolls down the track and then goes around a loop and when it leaves the loop flies off the inclined part of the track. ll the dimensions are: H = 80 m, r = 15 m, h = 10 m, ϴ = 30. e. What is the force applied by the surface on the car at point? Slide 74 / 76 F N + mg = mv 2 /r F N = mv 2 /r - mg F N = m(v 2 /r - g) F N = 900kg((39.6m/s) 2 /15m - 9.8m/s 2 ) F N = 49,961N kg roller coaster car starts from rest at point rolls down the track and then goes around a loop and when it leaves the loop flies off the inclined part of the track. ll the dimensions are: H = 80 m, r = 15 m, h = 10 m, ϴ = 30. Slide 75 / 76 f. How far from point will the car land? v0 = 37.0m/s y0 = 10m Y-direction voy = v0sinθ = 37m/s(sin30) = 18.5m/s y = y0 +voyt + ½gt 2 y = 10m +(18.5m/s)t + (-4.9m/s 2 )t 2 t = 4.3s X-direction vox = 37m/s(cos30) = 32.2m/s x = x0 +voxt + ½at 2 x = voxt x = (32.2m/s)(4.3s) x = 138.5m
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