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PhysicsW.E.P.Calculation of workEasy2 minQB

PhysicsEasyinteger

A time dependent force $F = 8 t$ acts on a particle of mass 1 kg . If the particle starts from rest, the work done by the force during first 1 second (in joule) is :

Answer:

Solution:

$a = \frac{F}{m} = \frac{8 t}{1} = 8 t$

$\frac{d v}{d t} = 8 t$

$v = \int_{0}^{1} 8 t d t = [4 t^{2}]_{0}^{1} = 4 (1)^{2} - 4 (0)^{2} = 4 m s^{- 1}$

$W = Δ K = \frac{1}{2} m v^{2}$

$W = \frac{1}{2} (1) (4^{2}) = \frac{1}{2} (1) (16) = 8 J$

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

⏱ 2mℹ️ Source: QB

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