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http://tuhsphysics.ttsd.k12.or.us/Tutorial/Lessons/Energy/Energy.htm 1. What is the potential energy of a 5.4 Kg shot put that is 12 m in the air? PE = mgh = (5.4 kg)(9.8 N/kg)(12 m) = 635 J (Table of contents) 2. What mass of water must you elevate a distance of 500 m to get a potential energy of 3.24 x 1011 J? PE = mgh 3.24 x 1011 J = m(9.8 N/kg)(500 m) h = 66122448.98 = 6.6x107kg (Table of contents) 3. To what height must a .5 Kg baseball rise to get a potential energy of 300 J? PE = mgh 300 J = (.5 kg)(9.8 N/kg)h h = 61.22 m = 61 m (Table of contents) 4. What is the kinetic energy of a .545 Kg baseball going 38 m/s? KE = 1/2mv2 KE = 1/2(.545 kg)(38)2 KE = 393.49 J = 393.5 J (Table of contents) 5. What speed must a 5 Kg hammer have to have a kinetic energy of 500 J? KE = 1/2mv2 500 J = 1/2(5 kg)v2 v = 14.14 = 14 m/s (Table of contents) 6. A speeding bullet with a speed of 582 m/s has a kinetic energy of 34,000 J. What is its mass? KE = 1/2mv2 34,000 J = 1/2m(582 m/s)2 m = 0.20075 = .201 kg (Table of contents) 7. It takes 500 N of force to drive a nail. A .69 Kg hammer going 5 m/s will drive it in what distance? Here you have kinetic energy turning into work: KE = 1/2mv2 W = Fd So they are equal: 1 /2mv2 = Fd 1 /2(.69 kg)(5 m/s)2 = (500 N)d d = .017 m (Table of contents) 8. A 1000 Kg car going 12 m/s is stopped in a distance of 34 m. What force stopped it? Again, kinetic energy is turning into work: KE = 1/2mv2 W = Fd So they are equal: 1 /2mv2 = Fd 1 /2(1000 kg)(12 m/s)2 = F(34 m) F = 2118 N (Table of contents) 9. A pile driver lifts a 45 Kg pile driving head a distance of 6.6 m above a piling. It drives the piling in a distance of .15 m. What force does it exert? Here, potential energy turns into work. PE = mgh W = Fd So they are equal: mgh = Fd So good so far, but the tricky thing is what you use for h and d: The total change in height of the pile driver is the 6.6 m it falls before striking the piling, PLUS the .15 m it pushes the piling in: h = 6.6 m + .15 m = 6.75 m And the distance d that the work is done on the piling over is only .15 m d = .15 m (That is, the pile driver only pushes on the piling for this distance. so mgh = Fd (45 kg)(9.80 N/kg)(6.75 m) = F(.15 m) F = 19845 N (Table of contents) 10. A .545 Kg pop fly has an upward velocity of 34 m/s. How high in the air will it go? This is a classic Conservation of energy problem: Total Energy Before Before: = = Total Energy After After: The ball at zero elevation, moving The ball having reached some maximum height upward at 34 m/s. It has kinetic energy h. It is at rest (top of flight). It has potential and nothing else. energy, and nothing else. 1 2 /2mv = mgh So 1 /2mv2 = mgh (We could at this point cancel m, but let's not) 1 /2(.545 kg)(34 m/s)2 = (.545 kg)(9.8 N/kg)h h = 59 m (Table of contents) 11. A 1 Kg ball falls 20 m. What is its speed when it hits the ground? This is the exact opposite of the previous problem: Total Energy Before Before: = = The ball at rest at a height of 20 m. It has potential energy and nothing else. mgh = So mgh = 1/2mv2 Let's cancel the darned m: mgh = 1/2mv2 gh = 1/2v2 (9.8 N/kg)(20 m) = 1/2v2 v = 19.8 m/s (Table of contents) Total Energy After After: The ball just before it hits the ground. It has kinetic energy and nothing else. 1 2 /2mv 12. A 1000 Kg car is going 12 m/s at the bottom of a 5 m tall hill. How fast is it going at the top? = = Total Energy Before Before: The car moving along at the top of the hill at some unknown velocity. It has both kinetic and potential energy. mgh + 1/2mvo2 = Total Energy After After: The car at the bottom of the hill moving at 12 m/s. It has kinetic energy, and nothing else. 1 2 /2mvf So mgh + 1/2mvo2 = 1/2mvf2 Notice that m will cancel at this point: gh + 1/2vo2 = 1/2vf2 (9.8 N/kg)(5 m) + 1/2vo2 = 1/2(12 m/s)2 vo = 6.78 m/s (Table of contents) 13. A 100 Kg rollercoaster is going 5 m/s at the top of a 7.5 m hill. What is its velocity at the top of a 5.0 m tall hill? Total Energy Before = Total Energy After After: Before: The rollercoaster moving along at the top of the first hill at 5 m/s and a height of 7.5 m. It has both kinetic and potential energy. mgho + 1/2mvo2 The rollercoaster moving along at the top of the second hill at an unknown velocity and a height of 5.0 m. It has both kinetic and potential energy. 1 2 = mghf + /2mvf = So mgho + 1/2mvo2 = mghf + 1/2mvf2 Again, the m will cancel from every term: gho + 1/2vo2 = ghf + 1/2vf2 (9.8 N/kg)(7.5 m) + 1/2(5 m/s)2 = (9.8 N/kg)(5.0 m) + 1/2vf2 vf = 8.6 m/s (Table of contents) 14. A 200 Kg rollercoaster is going 12 m/s at the top of a 3.0 m tall hill. What is its velocity at the top of a 7.0 m tall hill? Total Energy Before = Total Energy After After: Before: The rollercoaster moving along at the top of The rollercoaster moving along at the top the second hill at an unknown velocity and a of the first hill at 12 m/s and a height of 3.0 height of 7.0 m. It has both kinetic and m. It has both kinetic and potential energy. potential energy. mgho + 1/2mvo2 = mghf + 1/2mvf2 = So mgho + 1/2mvo2 = mghf + 1/2mvf2 Again, the m will cancel from every term: gho + 1/2vo2 = ghf + 1/2vf2 (9.8 N/kg)(3.0 m) + 1/2(12 m/s)2 = (9.8 N/kg)(7.0 m) + 1/2vf2 vf = 8.1 m/s (Table of contents) 15. A 300 Kg rollercoaster is going 5 m/s at the top of an 8 m tall hill. What is its speed at the bottom? Total Energy Before Before: = = The rollercoaster moving along at the top of the hill at 5 m/s and a height of 8 m. It has both kinetic and potential energy. mgho + 1/2mvo2 = So mgho + 1/2mvo2 = 1/2mvf2 Again, the m will cancel from every term: gho + 1/2vo2 = 1/2vf2 (9.8 N/kg)(8 m) + 1/2(5 m/s)2 = 1/2vf2 vf = 13.5 m/s (Table of contents) Total Energy After After: The rollercoaster at the bottom of the hill. It has only kinetic energy and nothing else. 1 2 /2mvf