Question

The gravitational intensity of a region is \[10{{ }}\left( {i - j} \right){{ }}N/kg\]. The work done by the gravitational force to shift slowly a particle of mass \[1{{ k}}g\] from point \[\left( {1m,{{ }}1m} \right)\] to a point \[\left( {2m, - 2m} \right)\] is-
A.\[10J\]
B.\[ - 10J\]
C.\[ - 40J\]
D.\[40J\]

Answer 1

Hint: From the given data, we will have to determine the gravitational force. Then, with the given initial and final positions vectors, we will be able to calculate the displacement. Work done by the gravitational force will be equal to the gravitational force into displacement.

Step by step answer:
Gravitational force: ${\vec F_g} = m{\vec E_g}$
Where ${\vec F_g}$ is the are the gravitational force and is expressed in Newton $(N)$, ${\vec E_g}$ is the intensity of the gravitational field and is expressed in Newton per kilograms $(N/kg)$ and $m$ is the mass of the body the gravitational field is acting on and is expressed in kilograms $(kg)$.
Displacement: $\vec s = \vec A - \vec B$
Where $\vec s$ is the displacement vector and is expressed in meters $(m)$, $\vec A$ is the initial position vector of the body and is expressed in meters $(m)$ and $\vec B$ is the final position vector of the body and is expressed in meters $(m)$.
Work done by the gravitational force: ${W_g} = {\vec F_g}\vec s$
Where $\vec s$ is the displacement vector and is expressed in meters $(m)$, $\vec A$ is the initial position vector of the body and is expressed in meters $(m)$ and $\vec B$ is the final position vector of the body and is expressed in meters $(m)$.

Given,
Intensity of gravitational field ${\vec E_g} = 10\left( {i - j} \right)N/kg$
Mass of body the above gravitational field is acting on $m = 1kg$
Initial position of body $\vec A = \left( {1m,{{ }}1m} \right)$
Final position of body $\vec B = \left( {2m, - 2m} \right)$
We need to find the gravitational force acting on the body in the field.
Therefore,
$
  {{\vec F}_g} = m{{\vec E}_g} = 1 \times 10\left( {i - j} \right)N/kg \\
   \Rightarrow {{\vec F}_g} = 10\left( {i - j} \right)N \\
 $
Now, to determine the displacement of the body we will subtract the initial position from the final position.
Therefore,
$
  \vec s = \vec B - \vec A = \left( {2m, - 2m} \right) - \left( {1m,{{ }}1m} \right) \\
   \Rightarrow \vec s = \left( {2\hat i, - 2\hat j} \right) - \left( {1\hat i,{{ }}1\hat j} \right) = (\hat i - 3\hat j) \\
   \Rightarrow \vec s = (\hat i - 3\hat j) \\
 $
Now that we know the displacement, we can determine the work done by the gravitational force on it on the body.
Therefore,
$
  {W_g} = {{\vec F}_g}\vec s = 10\left( {i - j} \right).(\hat i - 3\hat j) \\
   \Rightarrow {W_g} = 10(1 + 3) = + 40 \\
   \therefore {W_g} = + 40J \\
 $
Thus, the work done by the gravitational force to shift slowly a particle of mass \[1{{ k}}g\] from point \[\left( {1m,{{ }}1m} \right)\] to a point \[\left( {2m, - 2m} \right)\] is $40J$.

Hence, the correct option is D.

Note: Work is a scalar quantity because it is the dot product of displacement and force. Therefore, the similar unit vectors multiply to give one while the different ones give zero as a product.