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An object of mass 5 kg falls from rest through a vertical distance of 20 m and attains a velocity of 10 m/s. The work done by the resistance of air on the object is .....(g =10ms-2)

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Answered by kabir80
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Answered by TheUnsungWarrior
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Given: An object of mass 5 kg falls from rest through a vertical distance of 20 m and attains a velocity of 10 m/s.

To find: The work done by the resistance of air on the object.

Solution: The work done by the resistance of air will give out the work done by the frictional force. We know by work-energy theorem that work done by all the forces is equal to the change in the kinetic energy. So,

             Wgf + Wfr = Δ K.E.

where, Wgf = work done by gravitational force

            Wfr = work done by frictional force

So, putting the given values in the formula, we get:

           mgH + Wfr = 1/2 × m(Vf² - Vi²)

We took mgH to be +ve as displacement and gravity are acting in the same direction.

         5×10×20 + Wfr = 1/2 × 5 (10² - 0²)

           1000 + Wfr = 2.5 × 100

                       Wfr = 250 - 1000

                       Wfr = - 750 J

Hence, the work done by the resistance of air on the object is - 750 J.

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