Solveeit Logo
Login

Question

Question: A body is displaced from \[\left( {0,0} \right)\] to \[\left( {1\,{\text{m}},1\,{\text{m}}} \right)\...

A body is displaced from (0,0)\left( {0,0} \right) to (1 m,1 m)\left( {1\,{\text{m}},1\,{\text{m}}} \right) along the path x=yx = y by a force F=(x2j^+yi^) NF = \left( {{x^2}\hat j + y\hat i} \right)\,{\text{N}} . The work done by this force will be:
A. 43 J\dfrac{4}{3}\,{\text{J}}
B. 56 J\dfrac{5}{6}\,{\text{J}}
C. 32 J\dfrac{3}{2}\,{\text{J}}
D. 75 J\dfrac{7}{5}\,{\text{J}}

Explanation

Solution

First of all, we will re-arrange the quantities in the vector form of the force given. We will integrate it from the initial position to the final position and manipulate accordingly.

Complete step by step answer:
In the given question, we are supplied with the following data:
The initial position of the body is (0,0)\left( {0,0} \right) .
The final position of the body is (1 m,1 m)\left( {1\,{\text{m}},1\,{\text{m}}} \right) .
The body is moving in a straight line on the graph with the condition x=yx = y .
The magnitude of force is given by the vector form, F=(x2j^+yi^) NF = \left( {{x^2}\hat j + y\hat i} \right)\,{\text{N}} .
We are asked to find out the magnitude of work done in doing so.

Since, we are given, x=yx = y , we can write:
F = \left( {{x^2}\hat j + y\hat i} \right)\,{\text{N}} \\\
\implies F = \left( {{y^2}\hat j + x\hat i} \right)\,{\text{N}} \\\
We have just re-arranged the above expression of force.
We know, work done is the amount of energy spent in doing so and it can be written as the product of force and distance traversed.
W=∫F⃗.dr⃗W = \int {\vec F.d\vec r} …… (1)
Where,
F⃗\vec F indicates the amount of force acting on the body.
dr⃗d\vec r indicates the small distance traversed.
WW indicates the amount of work done.
The work can be assumed to be separated into two parts i.e. one part along the x-axis while the other part is along the y-axis.
So, to find the total work we will add both the components of work along the two axes.

Now, the equation (1), can be written as:
W = \int {{F_x}.dx} + \int {{F_y}.dy} \\\
\implies W = \int_0^1 {xdx} + \int_0^1 {{y^2}dy} \\\
\implies W = \left[ {\dfrac{{{x^2}}}{2}} \right]_0^1 + \left[ {\dfrac{{{y^3}}}{3}} \right]_0^1 \\\
\implies W = \left[ {\dfrac{{{1^2} - {0^2}}}{2}} \right] + \left[ {\dfrac{{{1^3} - {0^2}}}{3}} \right] \\\
Simplifying the above expression further, we get:
W = \dfrac{1}{2} + \dfrac{1}{3} \\\
\implies W = \dfrac{5}{6}\,{\text{J}} \\\
Hence, the required work done is found out to be 56 J\dfrac{5}{6}\,{\text{J}} .

So, the correct answer is “Option B”.

Note:
While doing this problem, you should have some knowledge of integration. It is important to take limits in the integration as this would have a definite answer. The limits so taken should be from00 to 11 . Remember that the work done will have two components as the force is inclined with an angle with horizontal.