Animated Solution for Physics - Work, Energy, and Power: A mass of $M$ kg is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of $45^{\circ }$ with the initial vertical direction is

Visualizing the Pendulum

Imagine a simple pendulum hanging straight down. Now, we apply a horizontal force $F$ to slowly push it aside until it makes a $45^{\circ }$ angle with the vertical. Our goal is to find the exact magnitude of this horizontal force $F$ in terms of the mass $M$ and gravity $g$.

To solve this, we need to look at the energy of the system. Since the mass is displaced slowly, it doesn't gain any speed. This means its initial and final kinetic energies are both zero.

Work-Energy Theorem

This is a perfect scenario for the Work-Energy Theorem, which states that the net work done by all forces on an object equals its change in kinetic energy:

Since the mass is displaced slowly, $\Delta K=0$. Therefore, the sum of the work done by all forces must be zero.

Let's identify the forces acting on the mass and calculate the work done by each:

1. Tension ($T$): The tension in the string always acts along the string, pointing towards the pivot. As the mass moves along the circular arc, its instantaneous displacement is always tangential to the arc. Since the radius (tension) and tangent (displacement) are always perpendicular, the work done by tension is zero: $W_T=0$.

2. Applied Horizontal Force ($F$): Work is defined as force multiplied by the displacement in the direction of the force. The force $F$ is purely horizontal. Using basic trigonometry, the horizontal displacement $x$ of the mass when it reaches an angle $\theta =45^{\circ }$ is $l\sin 45^{\circ }$. Thus, the work done by $F$ is positive:

3. Gravity ($Mg$): Gravity acts straight down. The mass, however, is moving upwards against gravity. The initial vertical depth of the mass is $l$. At $45^{\circ }$, its vertical depth is $l\cos 45^{\circ }$. The vertical height $y$ it has gained is the difference: $l-l\cos 45^{\circ }$. Because gravity and the vertical displacement are in opposite directions, the work done by gravity is negative:

Now, we plug these work values back into our Work-Energy equation:

Let's substitute the known values for $\sin 45^{\circ }$ and $\cos 45^{\circ }$, which are both $\frac{1}{\sqrt{2}}$:

Notice how the length of the string $l$ appears in every term? We can safely cancel it out. This tells us a beautiful physical truth: the required force doesn't depend on how long the string is!

To isolate $F$, we multiply the entire equation by $\sqrt{2}$:

And there we have it! The horizontal force required to slowly displace the pendulum to $45^{\circ }$ is exactly $Mg(\sqrt{2}-1)$.