Animated Solution for Physics - Work, Energy, and Power: Two persons $A$ and $B$ perform same amount of work in moving a body through a certain distance $d$ with application of forces acting at angle $45^{\circ }$ and $60^{\circ }$ with the direction of displacement respectively. The ratio of force applied by person $A$ to the force applied by person $B$ is $\frac{1}{\sqrt{x}}$. The value of $x$ is ....... .

Have you ever tried pulling a heavy suitcase at the airport? You might have noticed that pulling it with a longer handle (at a lower angle) feels different than pulling it with a shorter handle (at a steeper angle). This problem explores exactly that phenomenon! We are going to see how the angle of our pull affects the force required to do the same amount of work.

In physics, work isn't just about sweating and getting tired; it has a very precise mathematical definition. The work done $W$ by a constant force $\vec{F}$ causing a displacement $\vec{d}$ is given by the dot product of the two vectors:

Here, $\theta$ is the angle between the force and the displacement. Why the cosine? Because only the component of the force that acts along the direction of motion ($F\cos \theta$) actually contributes to moving the object. The perpendicular component just tries to lift the object off the ground, doing zero work in the horizontal direction.

Person A vs. Person B

The problem sets up a perfectly fair contest. Person A and Person B are doing the exact same amount of work ($W_A=W_B$) to move a body through the same distance $d$.

However, their techniques differ: Person A pulls at an angle ${\theta }_1=45^{\circ }$. Person B pulls at a steeper angle ${\theta }_2=60^{\circ }$.

Let's translate this physical situation into our master equation.

For Person A:

For Person B:

Since the work done is equal, we can equate the two expressions:

Finding the Ratio

Notice how the distance $d$ appears on both sides of the equation? Since it's the same distance for both, it beautifully cancels out. This tells us something profound: the ratio of the forces depends only on the angles, not on how far they pull the block!

Let's rearrange the equation to find the ratio of the force applied by A to the force applied by B ($\frac{F_A}{F_B}$):

Now, we bring in our standard trigonometric values. We know that $\cos (60^{\circ })=\frac{1}{2}$ and $\cos (45^{\circ })=\frac{1}{\sqrt{2}}$. Substituting these in:

To simplify this fraction, we multiply the numerator by the reciprocal of the denominator:

We can simplify this further by remembering that $2=\sqrt{2}\times \sqrt{2}$. Canceling one $\sqrt{2}$ from the top and bottom leaves us with:

The problem states that the ratio of the forces is $\frac{1}{\sqrt{x}}$. We have just rigorously calculated this ratio to be $\frac{1}{\sqrt{2}}$.

By simply comparing our result with the given format:

It is crystal clear that the value inside the square root must be 2.

And there we have it! A beautiful blend of physical intuition and trigonometric simplification.