Work, energy and power
Work
When a force causes an object to move, some energyThe capacity of a system to do work or the quantity required for mechanical work to take place. Measured in joules (J). For example, a man transfers 100 J of energy when moving a wheelbarrow. is transferred. We call this transfer of energy 'Work done' or just 'Work'.
The distance moved by the force is the same as the distanceNumerical description of how far apart two things are. For example, the distance from Edinburgh to Glasgow is approximately 50 miles. moved by the object.
The work done, Ew by a force, F moving through a distance, d is given by:
\({E_w} = Fd\)
Units are joules, newtons and metres.
If the force is overcoming frictional forces, all or some of the work done by the force is converted to heat energy.
Energy
The work done by the force may also be converted to kinetic energy or potential energy of the object.
When work is converted to different forms of energy you can:
- use the work done relationship to calculate energy gained or lost
- use energy relationships to calculate work done
Question
An light aircraft of mass 1200 kg starts from rest and accelerates along a straight horizontal runway.
The aircraft engine produces a constant thrust of 3400 N. A constant frictional force of 400 N acts on the aircraft.
The aircraft takes off when it reaches a speed of 35 ms-1.
State the unbalanced force acting on the aircraft.
Unbalanced force acting on the aircraft equals:
\((3400- 400)\)
Unbalanced force equals:
\(3000 N\)
Question
Now calculate the distance travelled by the aircraft from its starting point until it takes off.
To calculate the distance you first have to calculate kinetic energy since work done by the unbalanced force is converted to the kinetic energy of the aircraft.
The formula for calculating kinetic energy is:
\({E_k} = \frac{1}{2}m{v^2}\)
Where:
\({E_k} = Kinetic\,energy\)
\(m = mass\)
\(v = velocity\)
\({E_k} = \frac{1}{2}m{v^2}\)
\(m = 1200kg\)
\(v = 35m{s^{- 1}}\)
\({E_k} = \frac{1}{2}\times 1200\times {35^2}\)
\(= 7.35 \times {10^5} J\)
Now use kinetic energy = work done, to calculate distance.
\(F= 3000 N\)
\({E_k}= W = Fd\)
\(7.35\times {10^5} = 3000 \times d\)
\(d= 245 m\)
The aircraft travels 245 m from its starting point until it takes off.
You can also use energy relationships when energy is converted from one form to another.
This applies whether or not the force acts in a straight line.
This means you can use energy relationships in problems where you cannot use the equations of motion.
Work, energy and power in a pendulum
Question
A pendulum consists of a light string of length 1.2 m and a bob of mass 0.02 kg.
At its highest point the bob is a vertical distance of 80 mm higher than the lowest point of its swing.
State the forms of energy that the bob has at:
- the lowest point of its swing
- the highest point of its swing
- at the lowest point of its swing the bob has only kinetic energy
- at the highest point of its swing the bob has only potential energy
Question
Now calculate the maximum speed of the pendulum bob.
As the bob swings from its highest to its lowest point, potential energy is converted to kinetic energy.
\(m = 0.02kg\)
\(g = 9.8 m {s^{ - 2}}\)
\(h = 80mm = 0.08m\)
\({E_k} =\frac{1}{2}m{v^2} = mgh = {E_p}\)
\(\frac{1}{2} \times 0.02 \times {v^2} = 0.02 \times 9.8 \times 0.08\)
\({v^2} = 1.568\)
\(v = 1.2522 = 1.3 m{s^{ - 1}}\)
Maximum speed of bob = 1.3 ms-1
Question
Why might the maximum speed be less than calculated?
Energy losses due to friction may mean not all the gravitational potential energy is converted into kinetic energy 鈥 some may turn into heat energy.