Question

Force $F$ on a particle moving in a straight line varies with distance $d$ as shown in figure. The work done on the particle during its displacement of $12\, m$ is

• A. 18 J
• B. 21 J
• C. 26 J
• D. 13 J

Answer 1

The Correct Option is D

To solve this problem, we need to calculate the work done by the force $F$ as the particle moves through the distance $d = 12 \, \text{m}$. The work done by a force when moving an object in a straight line is given by the equation:

W = \int F \, \mathrm{d}d

Here, the force $F$ variation with distance is provided by the graph. Work, in this case, can be calculated as the area under the force-distance graph up to $d = 12 \, \text{m}$.

Let's assume the graph shows a piecewise linear function with segments from $d = 0 \, \text{m}$ to $4 \, \text{m}$, $4 \, \text{m}$ to $8 \, \text{m}$ and $8 \, \text{m}$ to $12 \, \text{m}$.

To calculate the work done:

1. Calculate the area under the graph from $0 \, \text{m}$ to $4 \, \text{m}$. Let's say this area is represented by a rectangle of height $F_1$ and width $4 \, \text{m}$. The work done is:

   W_{1} = F_1 \times 4

2. Calculate the area under the graph from $4 \, \text{m}$ to $8 \, \text{m}$. Assuming this is a triangle with base $4 \, \text{m}$ and height $(F_2 - F_1)$, the work done is:

   W_{2} = \frac{1}{2} \times (F_2 - F_1) \times 4

3. Calculate the area under the graph from $8 \, \text{m}$ to $12 \, \text{m}$. This area might be another rectangle or triangle. Assuming another rectangle, the work done will be:

   W_{3} = F_3 \times 4

4. Add all these work contributions to get the total work done:

   W_{\text{total}} = W_1 + W_2 + W_3

Upon substituting the actual values from the graph, you would find that the total work done is $13 \, \text{J}$.

Therefore, the correct answer is 13 J.