Question

If $\vec F = - {x^3}\hat i + {y^2}\hat j$ then find work done from $(2,3)$ to $(6,5)$ by using the Integration method.

Answer 1

Hint:In physics, work done by a body is the dot product between force vector and the displacement vector. In vector notation the displacement vector in XY plane can be written in form of $d\vec r = dx\hat i + dy\hat j$ and if vector force is denoted by $\vec F$ then according to the definition of work done we will use the formula for work done as $W = \int\limits_A^B {\vec F.d\vec r} $ where A,B are the initials and final points up to where body is displaced in XY plane.

Complete step by step answer:
 According to the question, we have given that, In X direction the body moves from point $x = 2$ to $x = 6$ and in Y direction the body is moved between points $y = 3$ to $y = 5$ now putting the values of $\vec F = - {x^3}\hat i + {y^2}\hat j$ and $d\vec r = dx\hat i + dy\hat j$ in the work done formula we have,
$W = \int\limits_{({x_1},{y_1})}^{({x_2},{y_2})} {\vec F.d\vec r} $
\[\Rightarrow W = \int\limits_{({x_1},{y_1})}^{({x_2},{y_2})} {( - {x^3}\hat i + {y^2}\hat j).(dx\hat i + dy\hat j)} \]
\[\Rightarrow W = \int\limits_{({x_1},{y_1})}^{({x_2},{y_2})} {( - {x^3}dx + {y^2}dy)} \]

Separating integration of variable x and variable y
\[W = \int\limits_{{x_1}}^{{x_2}} { - {x^3}dx + \int\limits_{{y_1}}^{{y_2}} {{y^2}dy} } \]
Since, using general integration formula we know that,
$\int {{a^n}dn = \dfrac{{{a^{n + 1}}}}{{n + 1}}} $ So we get,
\[\Rightarrow W = - [\dfrac{{{x^4}}}{4}]_{{x_1}}^{{x_2}} + [\dfrac{{{y^3}}}{3}]_{{y_1}}^{{y_2}}\]
Now, putting the limit values of x and y as
$x = 2$ To $x = 6$ and $y = 3$ to $y = 5$ we get,
\[W = - [\dfrac{{{x^4}}}{4}]_2^6 + [\dfrac{{{y^3}}}{3}]_3^5\]
On solving above equation we get,
\[W = - [\dfrac{{{6^4} - {2^4}}}{4}] + [\dfrac{{{5^3} - {3^3}}}{3}]\]
\[\Rightarrow W = - 320 + 32.66\]
\[\therefore W = - 287.34J\]

Hence, the work done by the body will be \[W = - 287.34\,J\].

Note:It should be remembered that, the SI unit of force is Newton and the SI unit of displacement is meter therefore the SI unit of work done is Newton-meter and $1Nm$ is written as Joule and the negative sign of work done indicates that work is done by the system and is work is done on the system then the sign of work done will be positive.