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)
Answers
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.