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Forces
Forces act on objects.
When a force causes an object to move, it is the force that does the work.
The distance moved by the force is the same as the distance moved by the object.
The work done by a force F moving through a distance d is given by:
W = Fd
If the force is overcoming frictional forces, all or some of the work done by the force is converted to heat 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, and energy relationships to calculate work done.
Remember energy cannot be created or destroyed - it is always converted to other forms of energy.
An aircraft of mass of 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 = (3400 - 400)
Unbalanced force = 3000 N
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} = ½mv^{2}
Where:
E_{k} = Kinetic energy
m = mass
v = velocity
E_{k }= ½mv^{2}
m = 1200 kg
v = 35 m s ^{-1}
E_{k} = ½× 1200 × 35^{2}
= 7.35 × 10^{5} J
Now use kinetic energy = work done to calculate distance.
F = 3000 N
E_{k }= W = Fd
7.35 × 10^{5} = 3000 × 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 moves in a straight line. This means you can use energy relationships in problems where you cannot use the equations of motion.
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 form of energy that the bob has at:
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 = 0.02 kg
g = 9.8 ms^{-2}
h = 80 mm = 0.08 m
E_{k} = ½mv^{2} = mgh = E_{p}
½ × 0.02 × v^{2} = 0.02 × 9.8 × 0.08
v^{2} = 1.568
v = 1.2522 = 1.3 ms^{-1}
Maximum speed of bob = 1.3 ms^{-1}
Power is the rate at which energy is used or work is done. This means that energy equals power × time, or
E = Pt.
This applies to all forms of work and all kinds of energy.
Note, in your examination you may be asked questions about electrical energy and electrical power.
A student of mass 64.0 kg runs up four flights of stairs, a total vertical height of 20.0 m, in 32.0 s.
Calculate the average power of the student.
First calculate the gain in gravitational potential energy
m = 64.0 kg
g = 9.8 N kg^{-1}
h = 20.0 m
Now use potential energy to calculate power.
t= 32.0 s
E = Pt
12544 = P × 32
P = 392 = 400 W
The power of the student is 400 W
Now state any assumption you have made in your calculation.
In practice, however, the student's power would be greater than the valued calculated since some energy is used in overcoming frictional forces.
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Links
Scottish Qualifications Authority Physics resources, including Past Papers and Arrangements Documents.
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