1. (part a) A bird (1.07 kg) glides horizontally (no wing flap) at 3 m/s. It the

Question

1. (part a) A bird (1.07 kg) glides horizontally (no wing flap) at 3 m/s. It then glides downward and then resumes horizontal gliding at a level 2 meters below its original gliding level. If it never flaps its wing, and we ignore any retarding air friction in the face of the bird, how fast should it be gliding at the of all of this?

(part b) Same as above, except that the bird uses its flapping wings as brakes during its decent. Now how fast should it be going at the end? Please show all calculations or reasoning.

Explanation / Answer

Solution:

Part a) Bird’s mass m = 1.07kg

Its initial velocity vi = 3m/s

Suppose it is gliding at some height h meters and take the gravitational potential of the bird Vi = m*g*h   Joule

(g = 9.8 m/s2 acceleration due to gravity )

The kinetic energy of the bird is given by Ki = (1/2)mvi2

Ki = (1/2)*(1.07kg)*(3m/s)2

Ki = 4.815 Joules

Thus total energy at height h is Ti = Ki + Vi

Ti = 4.185 J + m*g*h J

Since the bird do not flap, it means it does not apply breaking force and also retarding air friction is ignored. When the bird glides downwards it gravitational potential energy decreases and equal of energy is converted into its kinetic energy, thus its speed should increase.

The final height (h-2) m as bird descends 2 m; thus new final potential energy is

Vf = m*g*(h-2)   J

Vf = m*g*h – m*g*2

and the new kinetic energy is

Kf =(1/2)mvf2                            where vf is the final horizontal velocity of the bird

Thus the final total energy is

Tf = Kf + Vf

Tf = (1/2)mvf2 + (m*g*h – m*g*2)

Now since total energy should remain constant then it is means that

Ti = Tf

4.185 + m*g*h = (1/2)mvf2 + (m*g*h – m*g*2)

canceling mgh term on both sides, and rearranging we get

(1/2)mvf2 = 4.185 + m*g*2

vf2 = 2*(4.185 + m*g*2) / m

vf2 = 2*(4.185 + 1.07*9.8*2) / 1.07

vf2 = 47.02

taking square root we have,

vf = 6.86 m/s

Thus the new horizontal velocity of the bird is 6.86 m/s.

Part b)

The decrease in birds height is accompanied by increase in velocity. But now during the decent, the bird flaps its wing as a breaks. Thus although gravitational potential energy decreases, it doesn’t get converted into the kinetic energy due to breaking action. Thus horizontal speed of the bird should be same as before that is 3 m/s

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