Download 1. What is the potential energy of a 5.4 Kg shot put that is 12 m in the

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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
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
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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
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
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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
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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
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
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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
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
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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
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
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Total Energy After
After:
The rollercoaster at the bottom of the
hill. It has only kinetic energy and
nothing else.
1
2
/2mvf