Question:
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? with the initial vertical direction is :
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? with the initial vertical direction is :
Updated On: Jul 5, 2022
- $Mg\left(\sqrt{2}+1\right)$
- $Mg\sqrt{2}$
- $\frac{Mg}{\sqrt{2}}$
- $Mg \left(\sqrt{2}-1\right)$
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The Correct Option is D
Solution and Explanation
Here, the constant horizontal force required to take the body from position 1 to position 2 can be calculated by using work-energy theorem. Let us assume that body is taken slowly so that its speed doesn?t change, then $\Delta$K = 0
$=W_{F}+W_{M g}+W_{tension}$
[symbols have their usual meanings]
$W_{F}=F \times l \, sin \,45^{\circ}, $
$W_{Mg}=Mg \left(l-l \cos 45^{\circ}\right), w_{tension}=0$
$\therefore \, \quad F=Mg \left(\sqrt{2}-1\right)$
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Concepts Used:
Work-Energy Theorem
The work and kinetic energy principle (also known as the work-energy theorem) asserts that the work done by all forces acting on a particle equals the change in the particle's kinetic energy. By defining the work of the torque and rotational kinetic energy, this definition can be extended to rigid bodies.
The change in kinetic energy KE is equal to the work W done by the net force on a particle is given by,
W = ΔKE = ½ mv2f − ½ mv2i
Where,
vi → Speeds of the particle before the application of force
vf → Speeds of the particle after the application of force
m → Particle’s mass
Note: Energy and Momentum are related by, E = p2 / 2m.