A 10 - kg object is dropped from rest. after falling a distance of 50 m, it has a speed of 26 m/s. how much work is done by the dissipative (air) resistive force on the object during this descent?

Question

Jake Mueller

Physics

27 February, 02:42

A 10 - kg object is dropped from rest. after falling a distance of 50 m, it has a speed of 26 m/s. how much work is done by the dissipative (air) resistive force on the object during this descent?

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Answers (1)

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Kennedy Sparks · Answer 1

Using the equation of motion,

v^2 = u^2 + 2as, where v = final velocity, u = initial velocity = 0 m/s, a = acceleration = gravitational acceleration = 9.81 m/s^2, s = distance = 50 m

Therefore,

v = Sqrt (2*9.81*50) = 31.32 m/s

But, it is noted that the actual velocity is 26 m/s.

The difference between Kinetic Energies is the work done to overcome the air resistance.

That is,

Work done = 1/2*10*31.31^2 - 1/2*10*26^2 = 1/2*10 (31.32^2-26^2) = 1,525 J