Question

A 56 g Frisbee is thrown from a point 1.3 m above the ground with a speed of 15 m/s. When it has reached a height of 2.1 m, its speed is 12 m/s. What was the reduction in the mechanical energy of the Frisbee-Earth system because of air drag

Answer 1

The mechanical energy of frisbee-earth system because of air is equal to
`-2.2232J`

Step-by-step explanation:

From work energy theorem ;the total work done is equal to change in its kinetic energy

`w_(g)+w_(air)=K.E_(f)-K.E_(i)`

`w_(g)` = work done by gravity =
`mg(h_(1)-h_(2))`

`w_(air)` = work done by air drag

From above equations

`w_(air)` =
`(56)/(1000* 2 ) *(12^(2) -15^(2)) +(56)/(1000) *(2.1-1.3)J`

`w_(air)=-2.2232J`