under the action of a force, 2 kg body moves such that its position x as a function of time is given by x = (t)^(2/3) where x is in metres and t is in seconds. the work done by the force in the first two seconds is ?

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Vikas TU · Answer

velocity = dx/dt = (2/3)*t^-1/3 acceleration = (-2/9)*t^-4/3 Put t=2 in both v and accln. Work = Force x Velocity F=ma and velocity has been calculated already.

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under the action of a force, 2 kg body moves such that its position x as a function of time is given by x = (t)^(2/3) where x is in metres and t is in seconds. the work done by the force in the first two seconds is ?

under the action of a force, 2 kg body moves such that its position x as a function of time is given by x = (t)^(2/3) where x is in metres and t is in seconds. the work done by the force in the first two seconds is ?

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under the action of a force, 2 kg body moves such that its position x as a function of time is given by x = (t)^(2/3) where x is in metres and t is in seconds. the work done by the force in the first two seconds is ?

under the action of a force, 2 kg body moves such that its position x as a function of time is given by x = (t)^(2/3) where x is in metres and t is in seconds. the work done by the force in the first two seconds is ?

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