Question 180: Work, Energy and Power & Circular Motion

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 $${45^ \circ }$$ with the initial vertical direction is

Accepted Answer

From work energy theorem we can say,
Work done by tension $$+$$ work done by force (applied) $$+$$ Work done by gravitational force $$=$$  change in kinetic energy
Here Work done by tension is zero
$$ \Rightarrow 0 + F 	imes AB - Mg 	imes AC = 0$$
$$ \Rightarrow F = Mg\left( {{{AC} \over {AB}}} ight) = Mg\left[ {{{1 - {1 \over {\sqrt 2 }}} \over {{1 \over 2}}}} ight]$$
[ as $$AB = \ell \sin {45^ \circ } = {\ell  \over {\sqrt 2 }}$$
and $$AC = OC - OA = \ell  - \ell \,\cos \,{45^ \circ } = \ell \left( {1 - {1 \over {\sqrt 2 }}} ight)$$
where $$\ell  = $$ length of the string. ]
$$ \Rightarrow F = Mg\left( {\sqrt 2  - 1} ight)$$

Answer

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

Answer

$$Mg\sqrt 2 $$

Answer

$${{Mg} \over {\sqrt 2 }}$$

Answer

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

Work, Energy and Power & Circular Motion

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

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