Transcript Part33

Sample Problems for Part 3
on Conservation of Energy,
Power, and
Collisions/Explosions
Potential Energy #1
How much energy will it take to lift a 50 kg
object up 33 meters in height?
Since this is near the earth, we can use the
simpler form for gravitational
potential energy: PE = mgh.
h =33 m
The energy needed to lift is just
the change in PE, so
DPE = mghf – mghi = mg(Dh)
= (50 kg)*(9.8 m/s2)*(33 m)
= 16,170 Joules
f
hi=0 m
Potential Energy #2
Same problem of energy needed to lift an
object up, but this time we want to lift the
object up much higher – up to 330
kilometers.
Since we are no longer always close to the
earth’s surface, we will need to use the
more general form for gravitational
potential energy: PE = -G*M*m/r where r
is the distance from the center of M to the
center of m.
Potential Energy #2 – cont.
We again need to find the CHANGE in
potential energy:
DPE = PEf – PEi = (-GMm/rf) – (-GMm/ri) .
The radius in this case is the radius of the
earth plus the height, so
rf = Rearth + hf = 6,380 km + 330 km =
6,710 km = 6.71 x 106 m, and
ri = Rearth = hi = 6,380 km + 0 km =
6,380 km = 6.38 x 106 m.
Potential Energy #2 – cont.
DPE = PEf – PEi = (-GMm/rf) – (-GMm/ri)
rf = 6.71 x 106 m, and
ri = 6.38 x 106 m.
G = 6.67 x 10-11 Nt*m2/kg2
Mearth = 6.0 x 1024 kg
m = 50 kg
DPE = (-2.98 x 109 J) – (-3.14 x 109 J)
= 1.54 x 108 J .
Potential Energy #2 – cont.
Note:
If we had used the easier, but less accurate,
form for the potential energy due to gravity,
mgh, we would have obtained the less
accurate value of 1.62 x 108 J.
(instead of the more accurate value of 1.54 x 108 J)
Note that the more accurate value is less than the
constant gravity value, since the real gravity
decreases in strength as you move away from the
earth which lowers the energy needed to lift.
Potential Energy #2 – cont.
Note:
DPE = PEf – PEi = (-GMm/rf) – (-GMm/ri)
DOES NOT EQUAL (-GMm/([rf – ri]).
[You can see this from a simple example:
12/6 – 12/2 DOES NOT EQUAL 12/(6-2)
Since 2 – 6 DOES NOT EQUAL 3.]
Potential Energy & Power
A person on a bike with total mass of 75 kg
rides the bike up a hill that has a 4% grade
(which means that for every 100 meters of
road the height changes by 4 m). If the
person rides with a net power output of
250 watts, how fast can the rider go up the
hill?
Power = 250 watts
v=?
s
h = .04 s
Potential Energy & Power – cont.
Here we will neglect air resistance and friction. These
would either add to the required power or subtract from the
resulting speed.
Since power is the change in energy with
respect to time, and the energy change is
due to the increased height, we have:
Power = DPE / Dt = mg(Dh)/Dt .
Since we know Power, m, and g, we can solve
for (Dh/Dt):
Dh/Dt = Power / mg
= 250 watts / (75 kg * 9.8 m/s2) = 0.34 m/s .
Potential Energy & Power – cont.
Dh/Dt = 0.34 m/s
Due to the grade of 4%, Dh = 0.04*Ds, so
v = Ds/Dt = (Dh/0.04)/Dt = (0.34 m/s / 0.04)
= 8.5 m/s = 19 mph.
1-D Collisions
An object of mass 34 grams travelling South
with a speed of 122 m/s collides m=34 g
(and sticks to) an object of mass v=122 m/s
5,700 grams travelling North with
a speed of 2.6 m/s.
How fast will the combination of the
two masses be going, and in what v=2.6 m/s
direction?
m=5700 g
1-D Collisions – cont.
Since this is a collision problem, we should
think of Conservation of Energy and
Conservation of Momentum.
Since the two objects will end up sticking
together, v1f = v2f , and there will be some
energy lost.
Using the Conservation of Momentum:
m1v1i + m2v2i = m1v1f + m2v2f = (m1+m2)vf .
1-D Collisions – cont.
m1v1i + m2v2i = m1v1f + m2v2f = (m1+m2)vf .
We are given:
m1 = 34 g
v1i = 122 m/s South
m2 = 5,700 g
v2i = 2.6 m/s North
(34 g)*(122 m/s) + (5,700 g)*(-2.6 m/s)
= (34 g + 5,700 g)*vf , so
vf = -1.86 m/s .
Here we chose South as positive, so the negative
sign indicates the combination will be going
North.
1-D Collisions – cont.
Notes:
Minus signs mean something in physics.
Here be sure you assign positives to one
direction and negatives to the other.
I did not convert grams into kilograms,
although that is usually what I recommend.
Here the grams cancelled out so I didn’t
have to convert.
In space: throwing a light wrech or
heavy hammer
If an astronaut is floating in space with the safety
line cut, the astronaut can propel himself/herself
back towards the ship by throwing an object
away from the ship since by Newton’s 3rd law, if
the astronaut pushes an object one way, the
object will push the astronaut the other way.
Would a lighter wrench thrown faster work better
than throwing a heavier hammer slower?
v1f
v2f
In space: throwing a light wrech or
heavy hammer – cont.
Lets assume the light wrench has a mass of 0.1 kg
and the heavy hammer has a mass of 2 kg. The
astronaut plus tools (before the throw) has a total
mass of 75 kg.
Lets also assume that the light wrench can be thrown
at a speed of 30 m/s, while the heavy hammer can
be thrown at a speed of 5 m/s.
Which gives the bigger push to the astronaut?
Which takes more energy to throw?
In space: throwing a light wrech or
heavy hammer – cont.
Since this is an “explosion” problem, we
start with Conservation of Momentum:
m1v1i + m2v2i = m1v1f + m2v2f
Both initial speeds are zero: v1i = v2i = 0.
For the wrench: v1f = 30 m/s; m1 = 0.1 kg
m2 = 75 kg – 0.1 kg = 74.9 kg; v2f = ?
For the hammer: v1f = 5 m/s; m1 = 2.0 kg
m2 = 75 kg – 2.0 kg = 73 kg; v2f = ?
In space: throwing a light wrech or
heavy hammer – cont.
Solving for v2f using the wrench:
0 + 0 = (0.1 kg)(30 m/s) + (74.9 kg)v2f so
v2f = 3 kg*m/s / 74.9 kg = 0.040 m/s.
Solving for v2f using the hammer:
0 + 0 = (2.0 kg)(5 m/s) + (73.0 kg)v2f so
v2f = 10 kg*m/s / 73 kg = 0.137 m/s.
The hammer with 20 times more mass and only 6
times less speed gave the bigger kick to the
astronaut.
In space: throwing a light wrech or
heavy hammer – cont.
Which takes more energy to throw?
From the kinetic energy formula:
KE = ½ mv2
For the wrench:
KE = ½ (0.1 kg)(30 m/s)2 + ½ (74.9 kg)(.040 m/s)2 = 45.06 J.
For the hammer:
KE = ½ (2.0 kg)(5 m/s)2 + ½ (73.0 kg)(.137 m/s)2 = 25.68 J.