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Published byAina Frøydis Caspersen Modified over 6 years ago
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PE, KE Examples Answers 1. A shotput has a mass of 7.0 kg. Find the potential energy of a shotput raised to a height of 1.8 m. m = 7.0 kg h = 1.8 m PE = m g h = ( 7.0 kg )( 9.8 m/s2 )( 1.8 m ) PE = 123 J
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2. If a car of mass 1000 kg is raised to a height of 2.5 m
by a hydraulic lift, what is its PE? m = kg h = 2.5 m PE = m g h = ( 1000 kg )( 9.8 m/s2 )( 2.5 m ) PE = J
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3. A bowling ball of mass 7.0 kg moves at 6.5 m/s down a
lane. Find the kinetic energy of the ball. m = 7.0 kg v = 6.5 m/s KE = ½ m v2 = ½ ( 7.0 kg )( 6.5 m/s )2 KE = 148 J
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4. A car of mass 1200 kg moves at 24 m/s down a street.
Find the KE of the car. m = kg v = 24 m/s KE = ½ m v2 = ½ ( 1200 kg )( 24 m/s )2 KE = J KE = J
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5. A ball of mass 2.0 kg is raised to a height
of 2.5 m and held there. 2.0 kg (a) How much work was done to lift it to that height? 2.5 m W = F d F = force needed to lift ball = weight of ball = m g d = distance raised = height
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5. A ball of mass 2.0 kg is raised to a height
of 2.5 m and held there. 2.0 kg (a) How much work was done to lift it to that height? 2.5 m W = m g h F = force needed to lift ball = weight of ball = m g d = distance raised = height
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5. A ball of mass 2.0 kg is raised to a height
of 2.5 m and held there. 2.0 kg (a) How much work was done to lift it to that height? 2.5 m W = m g h = ( 2.0 kg )( 9.8 m/s2 )( 2.5 m ) W = 49 J
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5. A ball of mass 2.0 kg is raised to a height
of 2.5 m and held there. 2.0 kg (b) What is the PE of the ball at that height? PE = m g h 2.5 m = ( 2.0 kg )( 9.8 m/s2 )( 2.5 m ) PE = 49 J This is the same as (a); the answer to (b) could also be obtained by the Work-Energy Theorem (the energy acquired by the ball = the work done on it)
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5. A ball of mass 2.0 kg is raised to a height
of 2.5 m and held there. PE = 49 J 2.0 kg (c) What is the KE at that height? KE = ½ m v2 2.5 m
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5. A ball of mass 2.0 kg is raised to a height
of 2.5 m and held there. PE = 49 J 2.0 kg (c) What is the KE at that height? KE = ½ m v2 2.5 m v = 0 KE = 0
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5. A ball of mass 2.0 kg is raised to a height
of 2.5 m and held there. PE = 49 J KE = 0 2.0 kg (d) What is the Total Energy of the system? By Conservation of Energy, Total Energy = PE + KE 2.5 m
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5. A ball of mass 2.0 kg is raised to a height
of 2.5 m and held there. PE = 49 J KE = 0 2.0 kg (d) What is the Total Energy of the system? By Conservation of Energy, Total Energy = PE + KE 2.5 m = 49 J + 0 Total Energy = 49 J
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5. (e) The ball is released and falls to the ground. At the
instant before the ball strikes the ground, what is the PE of the ball? PE = m g h h = 0 PE = 0 2.0 kg
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5. (f) The ball is released and falls to the ground. At the
instant before the ball strikes the ground, what is the KE of the ball? Total Energy = PE + KE 49 J = KE PE = 0 KE = 49 J 2.0 kg Total Energy = 49 J from (d)
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5. (g) What is the speed of the ball as it strikes the ground?
2 ( KE = ½ m v2 ) 2 Solve for v 2 KE = m v2 m m 2 KE v2 = PE = 0 m KE = 49 J 2.0 kg 2 KE 2 ( 49 J ) v = = m 2.0 kg Challenge: Prove that J v = 7.0 m/s = m/s kg
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5. (h) As the ball is falling, PE is transformed into KE.
When the ball is 1.0 m above the ground, what is the PE? PE = m g h = ( 2.0 kg )( 9.8 m/s2 )( 1.0 m ) 2.0 kg PE = J 1.0 m
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5. (i) What is the ball’s KE when it is 1.0 m above the ground?
Total Energy = PE + KE PE = 19.6 J 49 J = J + KE 2.0 kg J J 1.0 m KE = J Total Energy = 49 J from (d)
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5. (j) What is the ball’s speed when it is 1.0 m above the ground?
2 ( KE = ½ m v2 ) 2 2 KE = m v2 PE = 19.6 J m m KE = 29.4 J 2.0 kg 2 KE 1.0 m v2 = m 2 KE 2 ( 29.4 J ) v = = m 2.0 kg v = 5.4 m/s
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6. A spring that has a spring constant of 45.0 N/m is
compressed 23.0 cm. (a) What force is required to compress the spring? F = k x k = N/m x = cm = m F = ( 45.0 N/m )( 0.23 m ) F = N
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6. A spring that has a spring constant of 45.0 N/m is
compressed 23.0 cm. (b) What is the elastic potential energy of the spring? PEe = ½ k x2 k = N/m x = cm = m PEe = ½ ( 45.0 N/m )( 0.23 m )2 PEe = J
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6. (c) A mass of 1.2 kg is placed on the spring and the
spring is released. What is the speed of the mass as it leaves the spring? PE = J When spring is released, PE of spring is transformed into KE of the mass 1.2 kg Mass leaves spring with 1.19 J of KE KE = ½ m v2 2 KE 2 ( 1.19 J ) v = = = v = 1.4 m/s m 1.2 kg
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7. Consider : (a) Find the PE at A PE = m g h
200 kg 50 m 30 m A B C (a) Find the PE at A PE = m g h = ( 200 kg )( 9.8 m/s2 )( 50 m ) PE = J
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(c) the Total Energy at A
7. Consider : 200 kg 50 m 30 m A B C (b) the KE at A v = 0 KE = 0 (c) the Total Energy at A Total Energy = PE + KE = J + 0 Total Energy = J
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7. Consider : (d) the PE at B h = 0 PE = 0 (e) the KE at B
200 kg 50 m 30 m A B C (d) the PE at B h = 0 PE = 0 (e) the KE at B Total Energy = PE + KE J = KE KE = J
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7. Consider : (f) the speed at B 2 KE 2 ( 98 000 J ) v = = m 200 kg
A B C (f) the speed at B 2 KE 2 ( J ) v = = m 200 kg v = m/s
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7. Consider : (g) the PE at C PE = m g h
200 kg 50 m 30 m A B C (g) the PE at C PE = m g h = ( 200 kg )( 9.8 m/s2 )( 30 m ) PE = J
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7. Consider : (h) the KE at C Total Energy = PE + KE
200 kg 50 m 30 m A B C (h) the KE at C Total Energy = PE + KE J = J + KE J J KE = J
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7. Consider : (i) the speed at C 2 KE 2 ( 39 200 J ) v = = m 200 kg
A B C (i) the speed at C 2 KE 2 ( J ) v = = m 200 kg v = m/s
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