A mass of M kg is suspended by a weightless string . The horizontal force that is required to displace it until the string making an angle of 45 with the initial vertical direction is

Option 1)

Mg\left ( \sqrt{2} -1\right )

Option 2)

Mg\left ( \sqrt{2} +1\right )

Option 3)

Mg\sqrt{2}

Option 4)

\frac{Mg}{\sqrt{2}}

Answers (1)

As we learnt in

Net work done by all the forces give the change in kinetic energy -

W=\frac{1}{2}mv^{2}-\frac{1}{2}mv{_{0}}^{2}

W= k_{f}-k_{i}

- wherein

m=mass \: of\: the\: body

v_{0}= initial\: velocity

v= final\: velocity

 

 

 

Height rises by the ball  =L-\frac{L}{\sqrt{}2}

Therefore conservation of energy =\frac{1}{2}mv^{2}=mg.(L-\frac{L}{\sqrt{}2})

Since kinetic energy change = Work done 

Therefore work done =F.\frac{L}{\sqrt{}2}=mg(L-\frac{L}{\sqrt{}2})

Therefore F=mg(\sqrt{}{2-1})

Correct answer is (1)

 

 


Option 1)

Mg\left ( \sqrt{2} -1\right )

This is the correct option.

Option 2)

Mg\left ( \sqrt{2} +1\right )

This is an incorrect option.

Option 3)

Mg\sqrt{2}

This is an incorrect option.

Option 4)

\frac{Mg}{\sqrt{2}}

This is an incorrect option.

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