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Physics - W.E.P. Question with Solution | TestHub

PhysicsW.E.P.Calculation of workMedium2 minQB

PhysicsMediumsingle choice

A body of mass m is lifted from ground by a force which varies with height ' h ' as $F = 2 (1 - bh) mg$ , here 'b' is constant. Work done by this force over the first half distance of the ascent of body :-

Options:

Answer:

Solution:

$F - mg = ma$

$2 mg (1 - bh) - mg = m \cdot v \cdot \frac{dv}{dh} mg \int_{0}^{h} [2 (1 - bh) - 1] dh = m \int_{0}^{v} v \cdot dv v^{2} = 2 g [h - bh^{2}]$

Velocity will be equal to zero at $h = \frac{1}{b}$

$W = \int_{0}^{\frac{1}{2} b} 2 mg (1 - bh) = 2 mg [\frac{3 b}{8 b}] = \frac{3 mg}{4 b}$

Stream:JEESubject:PhysicsTopic:W.E.P.Subtopic:Calculation of work

⏱ 2mℹ️ Source: QB

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