4.1-What is work study — sohailsunny7
What is work or transferring energy
Let’s talk about what is work. The weight-lifter shown in Figure has powerful muscles. They can provide the force needed to lift a large weight above her head — about 2m above the ground. The force exerted by the weight-lifter transfers energy from her to the weights. We know that the weights have gained energy because, when the athlete releases them, they come crashing down to the ground.
Figure It is hard work being a weight-lifter
As the athlete lifts the weights and transfers energy to them, we say that her lifting force is doing work. ‘Doing work’ is a way of transferring energy from one object to another. In fact, if you want to know the scientific meaning of the word ‘energy’, we have to say it is ‘that which is transferred when a force moves through a distance’. So work and energy are two closely linked concepts.
In physics, we often use an everyday word but with a special meaning. Work is an example of this. Table describes some situations which illustrate the meaning of doing work in physics.
It is important to appreciate that our bodies sometimes mislead us. If you hold a heavy weight above your head for some time, your muscles will get tired. However, you are not doing any work on the weights, because you are not transferring energy to the weights once they are above your head. Your muscles get tired because they are constantly relaxing and contracting, and this uses energy, but none of the energy is being transferred to the weights
Table The meaning of ‘doing work’ in physics.
Calculating work done
Because doing work defines what we mean by energy, we start this chapter by considering how to calculate work done. There is no doubt that you do work if you push a car along the road. A force transfers energy from you to the car. But how much work do you do? Figure shows the two factors involved:
■■ the size of the force F — the bigger the force, the greater the amount of work you do
■■ the distance s you push the car — the further you push it, the greater the amount of work done.
So, the bigger the force, and the further it moves, the greater the amount of work done
In the example shown in Figure,
F = 300N and s = 5.0m,
so:
work done W = F×s = 300×5.0 = 1500 J
Figure You have to do work to start the car moving.
Energy transferred
Doing work is a way of transferring energy. For both energy and work the correct SI unit is the joule (J). The amount of work done, calculated using W = F×s, shows the amount of energy transferred:
Newtons, metres and joules
From the equation W = F×s we can see how the unit of force (the newton), the unit of distance (the metre) and the unit of work or energy (the joule) are related.
1 joule = 1 newton × 1 metre
1J = 1Nm
he joule is defined as the amount of work done when a force of 1 newton moves a distance of 1 metre in the direction of the force. Since work done = energy transferred, it follows that a joule is also the amount of energy transferred when a force of 1 newton moves a distance of 1 metre in the direction of the force.
Force, distance and direction
It is important to appreciate that, for a force to do work, there must be movement in the direction of the force. Both the force F and the distance s moved in the direction of the force are vector quantities, so you should know that their directions are likely to be important. To illustrate this, we will consider three examples involving gravity (Figure). In the equation for work done, W = F×s , the distance moved s is thus the displacement in the direction of the force.
Suppose that the force F moves through a distance s which is at an angle θ to F, as shown in Figure . To determine the work done by the force, it is simplest to determine the component of F in the direction of s. This component is Fcos θ, and so we have:
work done = (Fcos θ)×s
or simply:
work done = Fs cos θ
Worked example 1 shows how to use this.
Doing work
You drop a stone weighing 5.0 N from the top of a 50 m high cliff. What is the work done by the force of gravity?
force on stone F = pull of gravity = weight of stone = 5.0 N vertically downwards
Distance moved by stone is s = 50 m vertically downwards.
Since F and s are in the same direction, there is no problem:
work done = F × s
= 5.0 × 50
= 250 J
Not doing work
A satellite orbits the Earth at a constant height and at a constant speed. The weight of the satellite at this height is 500 N. What is the work done by the force of gravity?
force on satellite F = pull of gravity = weight of satellite = 500 N towards centre of Earth
Distance moved by satellite towards centre of Earth (i.e. in the direction of force) is s = 0
The satellite remains at a constant distance from the Earth. It does not move in the direction of F. The work done by the Earth’s pull on the satellite is zero because
F = 500 N but s = 0:
work done = 500 × 0
= 0 J
Figure The work done by a force depends on the angle between the force and the distance it moves.
WORKED EXAMPLE
A man pulls a box along horizontal ground using a rope (Figure). The force provided by the rope is 200N, at an angle of 30° to the horizontal. Calculate the work done if the box moves 5.0m along the ground.
Step 1 Calculate the component of the force in the direction in which the box moves. This is the horizontal component of the force:
horizontal component of force = 200cos30° ≈ 173N
Hint: Fcosθ is the component of the force at an angle θ to the direction of motion.
Step 2 Now calculate the work done:
work done = force × distance moved
= 173 × 5.0 = 865J
A gas doing work
Gases exert pressure on the walls of their container. If a gas expands, the walls are pushed outwards — the gas has done work on its surroundings. In a steam engine, expanding steam pushes a piston to turn the engine, and in a car engine, the exploding mixture of fuel and air does the same thing, so this is an important situation.
Figure When a gas expands, it does work on its surroundings.
Figure shows a gas at pressure p inside a cylinder of cross-sectional area A. The cylinder is closed by a moveable piston. The gas pushes the piston a distance s. If we know the force F exerted by the gas on the piston, we can deduce an expression for the amount of work done by the gas.
From the definition of pressure (pressure = force /area ), the force exerted by the gas on the piston is given by:
force = pressure×area
F = p×A and the work done is force×displacement:
W = p×A×s
But the quantity A×s is the increase in volume of the gas;
that is, the shaded volume in Figure. We call this ΔV, where the Δ indicates that it is a change in V. Hence the work done by the gas in expanding is:
W = pΔV
Notice that we are assuming that the pressure p does not change as the gas expands. This will be true if the gas is expanding against the pressure of the atmosphere, which changes only very slowly.
- Read More:Unveiling the Secrets of Galactic Birth
- Read More:Why Are Planets Round
- Read More:The Bootstrap Paradox Simple Explanation
- Read More:What is Speed Easy Learning With Examples
- Read More:States of Matter | Gas Easy learning with Examples