Question

Calculate the net work done by the wings of a 0.300 kg hawk as he flies in a straight line at an angle of 35° above the horizontal from a height of 10.0 m to a height of 50.0 m. His starting speed is 6.00 m/s and his final speed is 10.0 m/s. The drag force is an average of 2.5 N during the flight

Answer 1

`W_(wing) = 301.5 J`

Step-by-step explanation:

As we know by work energy theorem that net work done by all forces is equal to change in kinetic energy

so here we can say

`W_(wing) + W_(gravity) + W_(drag) = (1)/(2)m(v_f^2 - v_i^2)`

now we know that when plane moves upwards then

`W_(gravity) = -mgh`

`W_(gravity) = 0.300(-9.8)(50 - 10)`

`W_(gravity) = -117.6 J`

Now work done by drag force is given as

`W_{drag = -F_(drag) (h_2 - h_1)/(sin\theta)`

`W_(drag) = -2.5((50- 10)/(sin35))`

`W_(drag) = -174.3 J`

now we have

`W_(wing) - 174.3 - 117.6 = (1)/(2)(0.300)(10^2 - 6^2)`

`W_(wing) = 301.5 J`