< Classical Mechanics

Lecture 5

Part 1

Part 2

Work, Energy and Power

Classical Mechanics

o Mechanical energy is just the total of all types of energy (kinetic, gravitational potential).Although there is no formal formula for mechanical energy, it could look something like:
- \(E_m = E_k + E_p\)
o A moving body has kinetic energy:
- \(E_k = \frac{1}{2} mv^2\) and
o Potential is stored energy that is able to do work later:
- \(\Delta\ E_p = mg\Delta\ h\)
o We can treat situations as being in a system. A system is any collection of objects in a particular place that we have defined. We usually do this so we can describe a problem as having particular limits (e.g. "the ball is rolling down the slope.")
o Systems come in three main types, depending on what is conserved (kept constant) throughout the situation.
- Isolated system: both Matter and Energy is conserved
- Closed system: Matter conserved
- Open system: neither Matter not Energy is conserved
o An isolated system simply means a situation where nothing, not even energy, can enter or leave. Whatever you started with you finish with.
- Since isolated systems can't lose or gain energy, we have to be careful that non- conservative forces (such as friction) do not happen.
- Also, if work is done on or by the system it will change the energy of the system. In that case it can not be an isolated system
- As long as we deal with isolated systems we know that the total mechanical energy before and after will be equal \((E_{m1} = E_{m2} \rightarrow E_{k1} + E_{p1} = E_{k2}+ E_{p2})\).
o A very important principle is that energy is not being created or changed, it is only changing forms or transferring from one object to another. This is known as the Law of Conservation of Energy.
A person is sitting on a toboggan at the top of a 23.7 \(m\) tall hill. If the person and toboggan have a total mass of 37.3 \(kg\) , determine how fast they will be going when they reach the bottom of the hill. Assume there is no friction.

1. 21.6 \(m/s\)
2. 21.6 \(km/h\)
3. 0 \(m/s\)
o Work is a transfer of energy.
\(W = \Delta\ E\)
o Work also happens when a force causes an object to move through a displacement
\(W = Fd cos \theta\ (Joules)\)
Work is a vector quantity.

1. True
2. False
You are pulling a box with a rope at a 30.0o angle from the ground. The box moves 12.7 \(m\) when you pull along the rope with a force of 76.0 \(N\). Determine how much work you did.

1. 965 \(J\)
2. 483 \(J\)
3. 836 \(J\)
o Power is the rate at which work is done, i.e how quickly energy is being used. The formula to determine power is simply the rate of change of work over time:\(P = \frac{W}{t} = \frac{\Delta\ E}{t}\) and measured in \(Watts (Joules/second\) or \(Newton\) \(metre/second)\). Since the work done is equal to the force applied over a displacement, \(W = Fd\), we can substitute to the above relationship and can be expressed as: \(P = \frac{W}{t} = \frac{Fd}{t} \rightarrow P = Fv\) in \(Newton\) \(metre/second\) or \(Watts\). This relationship can only be used in situations where the object has a constant velocity.
A 350 \(W\) motor is being used to lift shingles at a constant velocity to the top of a roof. If one pack of shingles has a 23 \(kg\) mass, determine the velocity that the pack will be raised at.

1. 1.21 \(m/s\)
2. 0.64 \(m/s\)
3. 0.32 \(m/s\)
Wille E. Coyote is trying to drop a 10 \(kg\) boulder off a 10 \(m\) high cliff to hit the Roadrunner eating a bowl of birdseed on the road below. At this particular location, gravity is 10 \(N/kg\). He wants to know the speed of the boulder at 3.0 \(m\) and 0 \(m\).

1. 7.7\(m/s\), 12.3 \(m/s\)
2. 7.7\(m/s\), 14.1 \(m/s\)
3. 11.8\(m/s\), 12.3 \(m/s\)
4. 11.8\(m/s\), 14.1 \(m/s\)

Part 1

Part 2

Lecture 6