The work done by the push of air on an object of mass 10 kg falling from rest through a vertical distance of 10 m is 500 J. Find the velocity of the object at the end of 10 m fall: {g = 10 m/s2).
Correct answer: D. 10 m/s
- A. 20 m/sec
- B. 12 m/sec
- C. 5 m/sec
- D. 10 m/s
Explanation
Given that, mass = 10kg. Vertical distance = 10m. Since starting from rest Initial velocity = 0 m/s. Work done = 500J. Using the Work Kinetic Energy Theorem, which states, Work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle. W = Kf-Ki. Since initial velocity is zero, final velocity will also be zero using the formula. K.E. = 1/2 m v². K.E. = 1/2 (10)(0). K.E. = 0. W = Kf-0. 500J = Kf-0. Kf = 500J. Final kinetic energy being 500J. K.Ef. = 1/2 m vf². 500= ½ (10) (vf)². v = "10m/s".
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The work-energy theorem states that the net work done on an object equals its change in kinetic energy. Questions involve positive and negative work, variable forces and force-displacement graphs, acceleration or deceleration, and the difference between this theorem and conservation of mechanical energy, which also requires attention to potential energy and non-conservative work.
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