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A bird is flying with a speed of 18.0 m/s over water when it accidentally drops a 2.00 kg fish. If the altitude of the bird is 5.40 m and friction is disregarded, what is the mechanical energy of the system, and what is the speed of the fish when it hits the water?

Respuesta :

skyluke89
1) What is the mechanical energy of the system?
(here I assume that "system" refers to the fish, because we don't know the mass of the bird).
The mechanical energy of the fish is conserved during the entire free fall, because there are no friction acting on it, therefore we can calculate it at the beginning of its free fall, when it is dropped by the bird. At this instant, the velocity of the fish is equal to the speed of the bird [tex](v_i=18 m/s)[/tex], and the distance from the ground is h=5.40 m. The mechanical energy is the sum of its kinetic energy and its potential energy:
[tex]E=K+U= \frac{1}{2}mv_i^2+mgh=[/tex]
[tex]= \frac{1}{2}(2 kg)(18 m/s)^2 +(2kg)(9.81 m/s^2)(5.40m)=430 J[/tex]

2) When the fish hits the water, its potential energy has become zero, because the height from the water is now zero: h=0. This means that the mechanical energy E (which is conserved) is now only kinetic energy. Therefore we can write:
[tex]E=K_f = \frac{1}{2}mv_f^2 [/tex]
And we can find the final speed:
[tex]v_f = \sqrt{ \frac{2E}{m} }= \sqrt{\frac{2\cdot 430 J}{2 kg}} = 20.7 m/s[/tex]
snehashish65

(a) The mechanical energy of the system is 429.84 J.

(b) The speed of the fish when it hits the water is 20.73 m/s.

Given data:

The speed of flying bird is, v = 18.0 m/s.

The mass of fish is, m = 2.00 kg.

The altitude of bird is, h = 5.40 m.

(a)

Since the fish is carried by the bird initially. So, considering them as a part of single system. The mechanical energy (ME) of system is given as,

[tex]\rm ME = \rm kinetic\;\rm energy + \rm potential \;\rm energy\\ME = \dfrac{1}{2}mv^{2} + mgh\\\\ME = \dfrac{1}{2} \times 2.00 \times 18.0^{2} + (2.00 \times 9.8 \times 54.0)\\\\ME= 429.84 \;\rm J[/tex]

Thus, the mechanical energy of the system is 429.84 J.

(b)

When the fish is dropped to water at ground, then entire mechanical energy is converted into the kinetic energy (or potential energy becomes zero). So,

mechanical energy = kinetic energy

[tex]ME = \dfrac{1}{2}mv^{2}[/tex]

Here, v is the speed of fish, with which it hits the water.

Solving as,

[tex]429.84 = \dfrac{1}{2} \times 2.00\times v^{2}\\v=\sqrt{429.84} \\v = 20.73 \;\rm m/s[/tex]

Thus, the speed of the fish when it hits the water is 20.73 m/s.

Learn more about the mechanical energy of system here:

https://brainly.com/question/17152889

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