Sample Problems for Part 3on 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.

The energy needed to lift is justthe change in PE, so PE = mghf – mghi = mg(h) = (50 kg)*(9.8 m/s2)*(33 m) = 16,170 Joules

hi=0 m

hf=33 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:PE = 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.

PE = 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

PE = (-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:

PE = 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 & PowerA 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?

s h = .04 sv = ?

Power = 250 watts

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 = PE / t = mg(h)/t .Since we know Power, m, and g, we can solve

for (h/t): h/t = Power / mg

= 250 watts / (75 kg * 9.8 m/s2) = 0.34 m/s .

Potential Energy & Power – cont.

h/t = 0.34 m/s

Due to the grade of 4%, h = 0.04*s, so

v = s/t = (h/0.04)/t = (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(and sticks to) an object of mass5,700 grams travelling North witha speed of 2.6 m/s.

How fast will the combination of the two masses be going, and in what direction?

m=34 g

m=5700 g

v=122 m/s

v=2.6 m/s

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.