In the previous section we discussed about the concept of forces. If you have not reviewed or do not know about the concept of force, you can click here. In this section we'll talk about the concept of work.
It is best to start with the definition of the work. In physics, "When a force moves an object in the direction of the force, work is said to be done." Works is a scalar quantity. It's unit is denoted by joule. Mathematically, it can be written as follows.
$$W=F\times d$$
If the force is applied in some angle then the formula for the work done is as follows,
$$W=Fcos\theta\times d$$
If our given system is a gas, the the work done is defined as,
$$W=p\times \Delta V$$
where, $p:$ pressure of the gas
$\Delta V:$ change in the volume
Next what if we perform an experiment where the force is not constant but varies for a given displacement to be covered(between $d_{1}$ and $d_{2}$). If we then want to find the total work done for this continuous graph as shown below we simply calculate the area under the curve.
$$W=\int_{d_{1}}^{d_{2}}F dx$$
Here's an easy to understand video tutorial on work.
$Ex.$ A man pushed a car for $10m$ on a straight road with an average force of $300N$ at an angle of $30^{o}$. Calculate the work done by the man.
$sol^{n}=2598.076Joules$
It is best to start with the definition of the work. In physics, "When a force moves an object in the direction of the force, work is said to be done." Works is a scalar quantity. It's unit is denoted by joule. Mathematically, it can be written as follows.
$$W=F\times d$$
If the force is applied in some angle then the formula for the work done is as follows,
$$W=Fcos\theta\times d$$
If our given system is a gas, the the work done is defined as,
$$W=p\times \Delta V$$
where, $p:$ pressure of the gas
$\Delta V:$ change in the volume
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$$W=\int_{d_{1}}^{d_{2}}F dx$$
Here's an easy to understand video tutorial on work.
$Ex.$ A man pushed a car for $10m$ on a straight road with an average force of $300N$ at an angle of $30^{o}$. Calculate the work done by the man.
$sol^{n}=2598.076Joules$
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