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 :

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) Mg(√2+1) (B) Mg√2 (C) (Mg/√2) (D) Mg (√2-1)

Tardigrade Admin · Accepted Answer

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

Δ

K = 0

= W_{F} + W_{M g} + W_{t e n s i o n}

[symbols have their usual meanings]

W_{F} = F \times l s in 4 5^{\circ},

W_{M g} = M g (l - l cos 4 5^{\circ}), w_{t e n s i o n} = 0

∴ F = M g (2 - 1)

Q. 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 :

$M g (2 + 1)$

$M g 2$

$\frac{M g}{2}$

$M g (2 - 1)$

Q. 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 :

Mg(2​+1)

Mg2​

2​Mg​

Mg(2​−1)

$M g (2 + 1)$

$M g 2$

$\frac{M g}{2}$

$M g (2 - 1)$