An airplane of mass 1.60 ✕ 104 kg is moving at 66.0 m/s. the pilot then increases the engine's thrust to 7.70 ✕ 104 n. the resistive force exerted by air on the airplane has a magnitude of 5.00 ✕ 104 n. (a) is the work done by the engine on the airplane equal to the change of the airplane's kinetic energy after it travels through some distance through the air? yes no correct: your answer is correct. is the mechanical energy conserved? explain your answer. energy is not conserved because there is friction. (b) find the speed of the airplane after it has traveled 5.00 ✕ 102 m. assume that the airplane is in level flight throughout the motion.

(a) No, because the mechanical energy is not conserved

Explanation:

The work-energy theorem states that the work done by the engine on the airplane is equal to the gain in kinetic energy of the plane:

(1)

However, this theorem is only valid if there are no non-conservative forces acting on the plane. However, in this case there is air resistance acting on the plane: this means that the work-energy theorem is no longer valid, because the mechanical energy is not conserved.

Therefore, eq. (1) can be rewritten as

which means that the work done by the engine (W) is used partially to increase the kinetic energy of the airplane () and part is lost because of the air resistance ().

(b) 77.8 m/s

First of all, we need to calculate the net force acting on the plane, which is equal to the difference between the thrust force and the air resistance:

Now we can calculate the acceleration of the plane, by using Newton's second law:

where m is the mass of the plane.

Finally, we can calculate the final speed of the plane by using the equation:

where

is the final velocity

is the initial velocity

is the acceleration

is the distance travelled

Solving for v, we find

the motion of a pendulum is a classic example of mechanical energy conservation. a pendulum consists of a mass (known as a bob ) attached by a string to a pivot point. as the pendulum moves it sweeps out a circular arc, moving back and forth in a periodic fashion.

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.