Energy Problems: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | Go up
 - by Chris Murray, 2002

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
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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
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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
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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
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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
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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
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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
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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
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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
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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	=	Total Energy After
Before:  The ball at zero elevation, moving upward at 34 m/s.  It has kinetic energy and nothing else.	=	After:  The ball having reached some maximum height h.  It is at rest (top of flight).  It has potential energy, and nothing else.
1/2mv2	=	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
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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	=	Total Energy After
Before:  The ball at rest at a height of 20 m.  It has potential energy and nothing else.	=	After:  The ball just before it hits the ground.  It has kinetic energy and nothing else.
mgh	=	1/2mv2

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
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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	=	Total Energy After
Before:  The car moving along at the top of the hill at some unknown velocity.  It has both kinetic and potential energy.	=	After:  The car at the bottom of the hill moving at 12 m/s.  It has kinetic energy, and nothing else.
mgh + 1/2mvo2	=	1/2mvf2

So 
mgh + ¹/₂mv_o² =  ¹/₂mv_f²
Notice that m will cancel at this point:
gh + ¹/₂v_o² =  ¹/₂v_f²
(9.8 N/kg)(5 m) + ¹/₂v_o² =  ¹/₂(12 m/s)²
v_o = 6.78 m/s
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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
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.	=	After:  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.
mgho + 1/2mvo2	=	mghf +  1/2mvf2

So 
mgh_o + ¹/₂mv_o² =  mgh_f +  ¹/₂mv_f²
Again, the m will cancel from every term:
gh_o + ¹/₂v_o² =  gh_f +  ¹/₂v_f²
(9.8 N/kg)(7.5 m) + ¹/₂(5 m/s)² =  (9.8 N/kg)(5.0 m) +  ¹/₂v_f²
v_f = 8.6 m/s
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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
Before:  The rollercoaster moving along at the top of the first hill at 12 m/s and a height of 3.0 m.   It has both kinetic and potential energy.	=	After:  The rollercoaster moving along at the top of the second hill at an unknown velocity and a height of 7.0 m.   It has both kinetic and potential energy.
mgho + 1/2mvo2	=	mghf +  1/2mvf2

So 
mgh_o + ¹/₂mv_o² =  mgh_f +  ¹/₂mv_f²
Again, the m will cancel from every term:
gh_o + ¹/₂v_o² =  gh_f +  ¹/₂v_f²
(9.8 N/kg)(3.0 m) + ¹/₂(12 m/s)² =  (9.8 N/kg)(7.0 m) +  ¹/₂v_f²
v_f = 8.1 m/s
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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	=	Total Energy After
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.	=	After:  The rollercoaster at the bottom of the hill.  It has only kinetic energy and nothing else.
mgho + 1/2mvo2	=	1/2mvf2

So 
mgh_o + ¹/₂mv_o² = ¹/₂mv_f²
Again, the m will cancel from every term:
gh_o + ¹/₂v_o² =  ¹/₂v_f²
(9.8 N/kg)(8 m) + ¹/₂(5 m/s)² =  ¹/₂v_f²
v_f = 13.5 m/s
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