Answer to Question #114808 in Classical Mechanics for Monica Kucher

Question

A 5.0 kg drone is flying perfectly horizontally above the earth’s surface at a height of 40 m and a speed of 15 m/s. It then drops down to a height of 25 m (still horizontally) and reduces its speed to 6 m/s. What is the amount of work that is done on the drone to change its mechanical energy in this way?

Answer 1

As per the question,

mass of the drone $(m) = 5kg$

Height of the drone $(h_1)=40m$

speed of the drone $(u)=15m/sec$

final height $(h_2)=25m$

final speed $(v)= 6m/sec$

So, work done $(w)= mgh_1- mgh_2+\dfrac{m\times u^2}{2}-\dfrac{mv^2}{2}$

$=(5\times 9.8\times 40 - 5\times 9.8\times25 +\dfrac{5\times 15^2}{2}-\dfrac{5\times 6^2}{2})J$

$=(735+472.5 )J$

$=1207.5J$