What is the eagles kinetic energy right before it catches the mouse?

The answers are

(A) kinetic energy = 4334 J

(B) centripetal acceleration =10.8 m/s².

(C) lift force = 82.6 N

Solution procedure:

A) Do an energy balance to find your kinetic energy (KE) at the bottom of the arc. At the bottom of the arc the eagle will have increased its kinetic energy by the amount of potential energy (PE) lost, minus the work done by the wings on the air:

KE before the dive: ½ m v² = (.5) (4) (30) = 1800 J

Change in potential energy (ΔPE) = m g Δh = (4) (9.81) (-100) = -3924 J

Work done: 1400 J

So, KE after the dive = KE before dive - ΔPE - 1400 J = 1800 + 3934 - 1400 = 4334 J

(B) Now just calculate the velocity squared corresponding to that kinetic energy, and divide by the radius to get the centripetal acceleration:

4334 J = ½ m v² = (.5) (4) v², therefore v² = 2167 m²/s²

Centripetal acceleration is given by a = v²/r, so a = 2167 / 200 = 10.8 m/s²

C) If the eagle is at exactly the bottom of the arc where the velocity has only a horizontal component, the centripetal acceleration keeping the eagle in the arc trajectory is vertical. This centripetal acceleration is added to the gravitational acceleration to produce a net vertical acceleration. The eagle's wings must produce this acceleration by generating a force using Newton's 2nd law: F = m a = (4) (9.81 + 10.8) = 82.6 N

Answers -->

(A) Kinetic energy = 3,200 J

(B) Centripetal acceleration = 8 m/s^2

(C) Lift at catch = 32 N * please read the note at the bottom for an explanation, it's important

How to find all of this....

(A) Eagle's kinetic energy

Given:

m = 4.0 kg

d = 100 m

Vi = 30 m/s

W = 1,400 J

Since we're given work, it can be assumed that we need to find a final velocity for the eagle which is picking up speed in its dive, not remaining a constant 30 m/s. Since work = force * distance, we have (and discounting that the bird will probably be diving at an angle, not a perfect 100 m... don't remind anyone, it'll get harder...)

1,400 J = F * (100 m)

We solve for force through simple division, giving

F = 14 N

Now we need to get the acceleration. Force = mass * acceleration, and we have a known mass:

14 N = (4.0 kg) * a

We can solve for acceleration through simple division again, getting

a = 3.5 m/s^2

Now we can solve for the final velocity, given initial velocity, distance, and acceleration:

Vf^2 = Vi^2 + 2ad

Vf = SQRT { (30 m/s)^2 + [2 * (3.5 m/s^2) * (100 m) ] }

Vf = SQRT { (900 m^2/s^2) + [ 700 m^2/s^2 ] }

Vf = SQRT { 1,600 m^2/s^2 }

Vf = 40 m/s

Finally, we can plug in our velocity and our known mass into Kinetic energy = one-half * mass * velocity-squared

KE = (0.5) * (4.0 kg) * (40 m/s)^2

KE = (2.0 kg) * (1,600 m^2/s^2)

KE = 3,200 J

Awesome.

(B) Eagle's centripetal acceleration

v = 40 m/s

r = 200 m

a = v^2 / r

a = (40 m/s)^2 / (200 m)

a = (1,600 m^2/s^2) / (200 m)

a = 8 m/s^2

Cool.

(C) Eagle's lift at catch

Are you just asking for force as lift? Using Newton's 2nd, known mass and acceleration we just calculated

F = (4.0 kg) * (8 m/s^2)

F = 32 N

But that's not actually how you solve for lift. Lift is found given the surface area of a wing, air density, and speed of air over the top and bottom surfaces of the wing, respectively -- Bernoulli's equation. This isn't lift, it's just a force number. But as the necessary information isn't given, I imagine that this is what you want.