A student drags a suitcase on a flat airport floor. If the force applied and the distance moved both double, how does the work done change?

The correct option is: Work done becomes four times greater.

Step-by-step Explanation:

1. Formula for Work Done:
Work done ($W$) by a force is given by the formula:

$W=F\times d\times \cos \left(\theta \right)$

where:
- $F$ is the applied force,
- $d$ is the distance moved, and
- $\theta$ is the angle between the direction of the force and the direction of motion.

Assuming the angle $\theta$ remains constant, work done is directly proportional to both the force and the distance:

$W\propto F\times d$

2. Initial Work Done:
Let the initial force be $F_1$ and the initial distance be $d_1$.
So, initial work done $W_1$ is:

$W_1=F_1\times d_1$

3. New Work Done:
According to the problem, both the force applied and the distance moved double.
New force, $F_2=2F_1$
New distance, $d_2=2d_1$

Now, calculate the new work done $W_2$:

$W_2=F_2\times d_2$

$W_2=\left(2F_1\right)\times \left(2d_1\right)$

$W_2=4\times (F_1\times d_1)$

$W_2=4W_1$

Conclusion:
The new work done becomes 4 times the initial work done (four times greater).