Define \[1\,J\] work ?
Answer
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Hint: Learn about work and work done by an object. Learn about the units related to work done to solve this problem. The SI unit of work or energy is \[joule\]. In the CGS system the unit of work or energy is \[erg\]. Work done by an object is given by the scalar product of the force applied and displacement of the object.
Formula used:
Work done by an object to move a certain displacement is given by,
\[W = \vec F.\vec S\]
where \[W\] is the work done by the object, \[\vec F\] is the net constant applied force and \[\vec S\] is the displacement of the object.
Complete step by step answer:
We know that work done by any object is the scalar product of the force applied on the object with the displacement of the object by the force applied. It can be written as, , \[W = \vec F.\vec S\] where \[W\] is the work done by the object, \[\vec F\] is the net constant applied force and \[\vec S\] is the displacement of the object.
For a variable force we can write that as, \[W = \int {\vec F(S).d\vec S} \] where \[\vec F(S)\] is the variable net applied force and \[\int {d\vec S} \] is the net displacement of the object.And \[dW = \vec F(S).d\vec S\] is the work done for \[d\vec S\] displacement of the object.
Now, we know that the SI unit of work done is \[joule\]. The SI unit of force is \[newton\] and SI unit of displacement is \[metre\]. So, if we put \[1N\] force and \[1m\] displacement in the equation for work done we have, \[1J = 1N \times 1m\]. So, from here we can say that \[1\,J\] of work done mean that if on a body \[1\,N\] of net constant force is applied to traverse a displacement of \[1\,m\] then the work done by the body is said to be, \[1\,J\].
Note: We can see that this definition is not applicable for a variable force as the value of force will change with gradual increase in displacement up to \[1\,m\]. Then the force applied at each point of the path will be different. But the integration \[W = \int {\vec F(S).d\vec S} \] over the path of \[\int {d\vec S} = 1\,m\] will be \[1\,J\].
Formula used:
Work done by an object to move a certain displacement is given by,
\[W = \vec F.\vec S\]
where \[W\] is the work done by the object, \[\vec F\] is the net constant applied force and \[\vec S\] is the displacement of the object.
Complete step by step answer:
We know that work done by any object is the scalar product of the force applied on the object with the displacement of the object by the force applied. It can be written as, , \[W = \vec F.\vec S\] where \[W\] is the work done by the object, \[\vec F\] is the net constant applied force and \[\vec S\] is the displacement of the object.
For a variable force we can write that as, \[W = \int {\vec F(S).d\vec S} \] where \[\vec F(S)\] is the variable net applied force and \[\int {d\vec S} \] is the net displacement of the object.And \[dW = \vec F(S).d\vec S\] is the work done for \[d\vec S\] displacement of the object.
Now, we know that the SI unit of work done is \[joule\]. The SI unit of force is \[newton\] and SI unit of displacement is \[metre\]. So, if we put \[1N\] force and \[1m\] displacement in the equation for work done we have, \[1J = 1N \times 1m\]. So, from here we can say that \[1\,J\] of work done mean that if on a body \[1\,N\] of net constant force is applied to traverse a displacement of \[1\,m\] then the work done by the body is said to be, \[1\,J\].
Note: We can see that this definition is not applicable for a variable force as the value of force will change with gradual increase in displacement up to \[1\,m\]. Then the force applied at each point of the path will be different. But the integration \[W = \int {\vec F(S).d\vec S} \] over the path of \[\int {d\vec S} = 1\,m\] will be \[1\,J\].
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