First slide

Work Energy Theorem

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 45⁰ with the initial vertical direction is

Moderate

A

$Mg (\sqrt{2} + 1)$
B

$Mg \sqrt{2}$
C

$M g / \sqrt{2}$
D

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

Solution

The situation is shown in fig. (9)
The constant force F required to take the body from position 1 to position 2 can be calculated by applying work-energy theorem. We assume that the body is taken slowly so that its speed doesn't change, i.e. $ΔK = 0$
$W_{F} + W_{Mg} + W_{tension} = 0 Hero W_{F} = F \times lsin 45^{\circ} W_{Mg} = Mg (l - lcos 45^{\circ}) W_{tension} = 0 F \times lsin 45^{\circ} + Mg (l - lcos 45^{\circ}) = 0 or F = Mg (\sqrt{2} - 1)$

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