
The gravitational field in a region is given by $\overrightarrow{E}=(2i+3j)\,{}^{N}/{}_{kg}$. Find the work done by the gravitational field when a particle of mass $1\,kg$ is moved on the line $3y+2x=5$ from $(1\,m,\,1\,m)$ to $(-2\,m,\,3\,m)$.
Answer
564.6k+ views
Hint: To get the solution, use the integral formula of work done where we can substitute the formula for force and then integrate the formula of work done on the line, which will give us the work done on the particle. If the formula is known, then it becomes a question of calculus, where we just need to find the integration of the equation.
Formula used:
$\begin{align}
& W=\int{\overrightarrow{F}.\overrightarrow{ds}} \\
& \overrightarrow{F}=m\overrightarrow{E} \\
\end{align}$
Complete Step-by-Step solution:
In the question, the values given are:
$\overrightarrow{E}=(2i+3j)\,{}^{N}/{}_{kg}$,
Mass, $m=1\,kg$, and
line $3y+2x=5$ from $(1\,m,\,1\,m)$ to $(-2\,m,\,3\,m)$.
Now,
The work done by the gravitational field, can be found using the below formula:
$W=\int{\overrightarrow{F}.\overrightarrow{ds}}$
Here, we know that:
$\overrightarrow{F}=m\overrightarrow{E}$ and also we know the values of $\overrightarrow{E}$ and $m$, substituting the value in the work done formula we get:
$\begin{align}
& W=\int{m\overrightarrow{E}.\overrightarrow{ds}} \\
& \Rightarrow W=1.\int{\overrightarrow{E}.(dx\hat{i}+dy\hat{j})} \\
& \Rightarrow W=\int{{{E}_{x}}.dx}+\int{{{E}_{y}}.dy} \\
\end{align}$
Now, ${{E}_{x}}$ and ${{E}_{y}}$ can be found using the equation of the line.
$\begin{align}
& W=\int_{1}^{-2}{\left( \dfrac{2}{5} \right).dx}+\int_{1}^{3}{\left( \dfrac{3}{5} \right).dy} \\
& \Rightarrow W=\left( \dfrac{2}{5} \right)(0)+\left( \dfrac{3}{5} \right)(0) \\
& \therefore W=0 \\
\end{align}$
Therefore, no work is done by the gravitational field when the particle is moved on the line.
Note:
This question is related to calculus if the formula for work done is known. Calculus plays an important role in physics as well, as many questions in physics can be solved using the concept of calculus. If the formula is known, then finding the solution becomes easy, just the concept should be clear. This is not the only way to solve this problem, there are other approaches as well, but this is an efficient approach to get the answer.
Formula used:
$\begin{align}
& W=\int{\overrightarrow{F}.\overrightarrow{ds}} \\
& \overrightarrow{F}=m\overrightarrow{E} \\
\end{align}$
Complete Step-by-Step solution:
In the question, the values given are:
$\overrightarrow{E}=(2i+3j)\,{}^{N}/{}_{kg}$,
Mass, $m=1\,kg$, and
line $3y+2x=5$ from $(1\,m,\,1\,m)$ to $(-2\,m,\,3\,m)$.
Now,
The work done by the gravitational field, can be found using the below formula:
$W=\int{\overrightarrow{F}.\overrightarrow{ds}}$
Here, we know that:
$\overrightarrow{F}=m\overrightarrow{E}$ and also we know the values of $\overrightarrow{E}$ and $m$, substituting the value in the work done formula we get:
$\begin{align}
& W=\int{m\overrightarrow{E}.\overrightarrow{ds}} \\
& \Rightarrow W=1.\int{\overrightarrow{E}.(dx\hat{i}+dy\hat{j})} \\
& \Rightarrow W=\int{{{E}_{x}}.dx}+\int{{{E}_{y}}.dy} \\
\end{align}$
Now, ${{E}_{x}}$ and ${{E}_{y}}$ can be found using the equation of the line.
$\begin{align}
& W=\int_{1}^{-2}{\left( \dfrac{2}{5} \right).dx}+\int_{1}^{3}{\left( \dfrac{3}{5} \right).dy} \\
& \Rightarrow W=\left( \dfrac{2}{5} \right)(0)+\left( \dfrac{3}{5} \right)(0) \\
& \therefore W=0 \\
\end{align}$
Therefore, no work is done by the gravitational field when the particle is moved on the line.
Note:
This question is related to calculus if the formula for work done is known. Calculus plays an important role in physics as well, as many questions in physics can be solved using the concept of calculus. If the formula is known, then finding the solution becomes easy, just the concept should be clear. This is not the only way to solve this problem, there are other approaches as well, but this is an efficient approach to get the answer.
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