A body of mass 3 k g acted upon by a constant force is displaced by S metre, given by relation S = 1 3 t 2 , where t is in second. Work done by the force in 2 seconds is :

S = t 2 3 v = d S d t d S d t = 2 3 t = v ; W = ∆ K = K t = 2 - K t=0 W= 1 2 m v t = 2 2 - v t = 0 2 = 1 2 ⋅ 3 16 9 - 0 = 8 / 3 J

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

PhysicsW.E.P.MiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

A body of mass $3 k g$ acted upon by a constant force is displaced by $S$ metre, given by relation $S = \frac{1}{3} t^{2}$ , where $t$ is in second. Work done by the force in $2$ seconds is :

Options:

Answer:

Solution:

$S = \frac{t^{2}}{3}$

$v = \frac{d S}{d t}$

$\frac{d S}{d t} = \frac{2}{3} t = v;$

$W = ∆ K = K_{t = 2} - K_{t=0}$

$W= \frac{1}{2} m (v_{t = 2}^{2} - v_{t = 0}^{2}) = \frac{1}{2} \cdot 3 (\frac{16}{9} - 0) = 8 / 3 J$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:W.E.P.Subtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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