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Have you ever pushed someone on a swing, or thrown a ball? Have you ever pulled a suitcase along, or lifted a weight at the gym? These are just a few of countless examples of times when you might have done work on another object. Every day you do work constantly; not school work, or work at your job, but a special kind of work that is central to the study of energy transfer in physics. So, what exactly is work? It's all to do with forces.
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Team Work Done by a Constant Force Teachers
Work is simply the amount of energy transferred from one object to another by the application of unbalanced force. When you push a box along the floor, there are two horizontal forces at work: friction, and the force you yourself are exerting. If the force you exert on the box is greater than the force of friction, the box will move.
In other words, it will gain kinetic energy. How much kinetic energy? Well, that's dependent on how much work is done on the box; in fact, it is equal to how much work is done on the box.
\[E_k = W\]
The next question you may ask, is how much work is done by you exerting the force? Well, that is dependent on two factors: the magnitude of the unbalanced force, and the distance the box is pushed. The most simple case is where the box is pushed with a constant force. The formula for work done by a constant force is
\[W = Fd.\]
Where work, \(W\), is in joules (J), force, \(F\), is in newtons (N), and distance, \(d\), is in metres (m).
Work done by a constant force is the energy transferred to an object due to a force being exerted on it by another object.
For every metre an object is moved by a constant force of one newton, it will gain \(1\, J\) of kinetic energy. In fact, what you may not have realised is that the unit of joules is really just another term for newton-metres!
It's all very well stating this formula for work done by a constant force, but how does it come about? Let's have a go at deriving it.
Deriving any formula can seem intimidating, but rest assured, this one's not so bad! To do this, just think about how much energy the object has before the constant force is applied, \(t_1\) and some arbitrary amount of time later, \(t_2\). Well, the only energy change in the object that really happens is it gains kinetic energy. So the total amount of energy it has gained between \(t_1\) and \(t_2\) is just the difference in its kinetic energy at these times.
\[W = E_{k2} - E_{k1}\]
Remember the formula for kinetic energy?
\[E_k = \frac{1}{2}mv^2\]
Let's plug that into the formula above.
\[W = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 \]
Remember, the mass of the object stays the same the whole time, but the velocity is different at \(t_1\) and \(t_2\).
Now, do a bit of factorising.
\[W = \frac{1}{2}m (v^2-u^2)\]
If you think back to your equations of motion, you may recall that
\[v^2 - u^2 = 2ad\]
so, this can now be substituted into the formula.
\[\begin{align} W &= \frac{1}{2}m\times 2as \\ W&= mad \end{align}\]
Finally, remembering Newton's second law,
\[F = ma,\]
one more substitution can be made to arrive at the original formula for work done by a constant force.
\[W= Fd \]
There you have it! A nice quick derivation.
So what does this equation actually tell us? Well, all it says is that for every metre an object is pushed (or pulled) with a constant force of one newton, it will gain one joule of energy.
\[W = Fd\]
If you pushed a \(4\,\text{kg}\) box with a constant force of \(3\,\text{N}\) for \(10\,\text{m}\) then...
\[\begin{align} W &= 3\times 10 \\ &= 30\,\text{J} \end{align}\]
How fast is the box moving by the end of the \(10\,\text{m}\)? You can work that out too.
Remember that the kinetic energy gained by the object is equal to the total work done on the object, so...
\[\begin{align} E_k = W &= \frac{1}{2}mv^2 \\ 30 &= \frac{1}{2}\times 4v^2 \\ 30 &= 2v^2 \\ v^2 &= 15 \\ v &= 3.87\, \text{ms}^{-1} \end{align}\]
Let's take a look at some more examples to get to grips with this.
(1)
A bike with mass \(50\,\text{kg}\) rides down a hill from rest with an average force down the hill of \(170\,\text{N}\). The average force of friction experienced by the bike is \(30\,\text{N}\). Given that its \(30\,\text{m}\) from the top of the hill to the bottom, what is the speed of the bike at the bottom of the hill?
Solution:
First, calculate the work done on the bike going down the hill.
\[\begin{align} W &= Fd \\ &= 170 \times 30 \\ &= 5100\,\text{J} \end{align}\]
Then, calculate the opposing work done by friction up the slope.
\[\begin{align} W_f &= 30\times 30 \\ &= 900\,\text{J} \end{align}\]
Since the work done by friction is opposite to the work on the bike down the hill, the net work done is
\[\begin{align} 5100-900 = 4200\,\text{J}. \end{align}\]
Now, calculate the velocity of the bike at the bottom of the hill.
\[\begin{align} E_k = W &= \frac{1}{2}mv^2 \\ 4200 &= \frac{1}{2}\times 50v^2 \\ v^2 &= 84 \\ v &= 9.17\,\text{ms}^{-1} \end{align}\]
(2)
A car of mass \(2300\,\text{kg}\) drives up a hill with an average force of \(6000\,\text{N}\). The average force of the car's weight down the hill is \(4000\,\text{N}\), and the average force of friction is \(1800\,\text{N}\). Given that hill is \(60\,m\) from bottom to top, what is the speed of the car when it reaches the top?
Solution:
First, calculate the total work done by the car up the slope.
\[\begin{align} W &= Fd \\ &= 6000\times 60 \\ &= 360000\,\text{J} \end{align}\]
Next, calculate the work done on the car down the slope by its weight and friction.
\[\begin{align} W &= Fd \\ &= (4000+1800) \times 60 \\ &= 348000\,\text{J} \end{align}\]
Since the work done by friction and weight down the slope is opposite to the work on the bike down the hill, the net work done is
\[360000-348000 = 12000\,\text{J}.\]
Now, calculate the velocity of the car at the top of the hill.
\[\begin{align} W = E_k &= \frac{1}{2}mv^2 \\ 12000 &= \frac{1}{2}\times 2300v^2 \\ 12000&= 1150v^2 \\ v^2 &= 10.43 \\ v &= 3.23\,\text{ms}^{-1} \end{align}\]
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What is meant by a constant force?
A force that does not vary over time.
What is work done by a constant force when it is positive or negative?
When positive work is done by a constant force, the object gains kinetic energy in the direction of its motion. When negative work is done by a constant force, the object loses kinetic energy.
Which is the example of constant force?
The force of weight you experience is constant, though it does decrease the further you are from sea level.
What is the formula for finding the work done by a constant force in the same direction as the displacement?
W = Fd.
How will you find the work done by a variable force explain briefly?
Work done by a variable force can be found using calculus.
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