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

PhysicsW.E.P.MiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

Under the action of a force a $2 kg$ body moves such that its position $x$ in meters as a function of time $t$ is given by $x = \frac{t^{4}}{4} + 3$ . Then work done by the force in first two seconds is

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Answer:

Solution:

$x = \frac{t^{4}}{4} + 3$

$\frac{d x}{d t} = \frac{4 t^{3}}{4} + 0$

$v = t^{3}$

Using work energy theorem

$w = ∆ k = \frac{1}{2} m {(v (2))}^{2} - \frac{1}{2} m {(v (0))}^{2}$

$= \frac{1}{2} (2) [{(2^{3})}^{2} - 0^{3}]$

$= 64 J$

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

⏱ 2mℹ️ Source: PYQ_2020

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