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Step-by-Step Solution
Step 1: Understand the Physical Situation
A mass $M$ is hanging vertically at rest from a weightless string. We want to apply a constant horizontal force $F$ to move the mass from the initial vertical position until the string makes a $45^\circ$ angle with the vertical. Because the force is constant and acts horizontally, the mass will rise in height, thereby gaining potential energy.
Step 2: Use Energy Conservation
Since the only work done against gravity is by the net horizontal force (together with the tension), we can equate the work done by this constant horizontal force to the gain in gravitational potential energy. Let the length of the string be $L$.
Vertical displacement ($\Delta h$):
Initially, the mass is vertically below the point of support (string fully vertical).
Finally, when the string makes angle $45^\circ$ with the vertical, the mass is at a certain height above the original position.
The new vertical distance from the pivot is $L \cos 45^\circ = L \frac{1}{\sqrt{2}}$.
So,
$$
\Delta h \;=\; L - L \cos 45^\circ \;=\; L \bigl(1 - \frac{1}{\sqrt{2}}\bigr).
$$
Horizontal displacement ($x$):
The mass also moves horizontally from the original line by $L \sin 45^\circ = L \frac{1}{\sqrt{2}}.$
Gain in potential energy:
$$
\text{Gain in P.E.} = M g \,\Delta h = M g\, L \bigl(1 - \frac{1}{\sqrt{2}}\bigr).
$$
Work done by the constant horizontal force $F$:
$$
W = F \times (\text{horizontal displacement}) = F \times L \frac{1}{\sqrt{2}}.
$$
Step 3: Equate Work Done to Gain in Potential Energy
By energy conservation, the work done by the horizontal force must equal the increase in potential energy:
$$
F \times L \frac{1}{\sqrt{2}} \;=\; M g \, L \Bigl(1 - \frac{1}{\sqrt{2}}\Bigr).
$$
Canceling $L$ from both sides and rearranging for $F$ gives:
$$
F \;=\; M g \,\frac{1 - \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}
\;=\; M g \bigl(\sqrt{2} - 1\bigr).
$$
Step 4: Final Answer
Therefore, the constant horizontal force required is:
$$
\boxed{F = M g \bigl(\sqrt{2} - 1\bigr).}
$$
