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Question

Physics Question on Work-energy theorem

An object of mass 10kg10\, kg falls from rest through a vertical distance of 10m10\,m and acquires a velocity of 10m/s10\,m/s .The work done by the push of air on the object is (g=10m/s2)(g = 10 \, m/s^2)

A

500 J

B

- 500 J

C

250 J

D

- 250 J

Answer

- 500 J

Explanation

Solution

Given, mass of object =10kg=10\, kg
Initial velocity (rest), u=0u =0
Final velocity, v=10m/sv =10 m / s
g=10m/s2g =10\, m / s ^{2}
Work done by air =(=-( Work done by the object to cover the vertical distance)
=F×s=mgh[as,v2u2=2ghgh=v2u22]=- F \times s =- mgh [as, v ^{2}- u ^{2}=2 gh \Rightarrow gh =\frac{ v ^{2}- u ^{2}}{2}]
=10×(10202)2=10×10×102=500J=-10 \times \frac{\left(10^{2}-0^{2}\right)}{2}=-\frac{10 \times 10 \times 10}{2}=-500\, J