An airplane of mass 1.5x10^4kg is moving at 60m/s. The pilot then revs up the engine so that the forward thrust of the propeller becomes 7.5x10^4 N, find the speed of airplane after it has traveled 500m. Assume that the airplane is in level flight through this motion.

To find the final speed of the airplane after it has traveled 500m, we can use the principle of conservation of energy and apply Newton's second law of motion.

1. Calculate the initial kinetic energy (K1) of the airplane using the mass (m) and initial speed (v1) formula:
K1 = 0.5 * m * v1^2

With mass (m) = 1.5 x 10^4 kg and initial speed (v1) = 60 m/s:
K1 = 0.5 * 1.5 x 10^4 kg * (60 m/s)^2

2. Calculate the work done on the airplane (W) by the forward thrust force (F) over a distance (d) formula:
W = F * d

With forward thrust force (F) = 7.5 x 10^4 N and distance (d) = 500 m:
W = 7.5 x 10^4 N * 500 m

3. Apply the work-energy principle, which states that the work (W) done on an object is equal to the change in its kinetic energy (Delta K):
W = Delta K

Delta K = K2 - K1, where K2 is the final kinetic energy of the airplane.

W = K2 - K1

4. Rearrange the equation to solve for the final kinetic energy (K2):
K2 = W + K1

5. Calculate the final speed (v2) of the airplane using the mass (m) and final kinetic energy (K2) formula:
K2 = 0.5 * m * v2^2

Rearrange the equation to solve for the final speed (v2):
v2 = sqrt(2 * K2 / m)

Now, let's substitute the given values into the formulas and solve step-by-step:

Step 1:
K1 = 0.5 * 1.5 x 10^4 kg * (60 m/s)^2
= 54000000 J

Step 2:
W = 7.5 x 10^4 N * 500 m
= 37500000 J

Step 3:
W = K2 - K1
37500000 J = K2 - 54000000 J
K2 = 37500000 J + 54000000 J
= 91500000 J

Step 4:
v2 = sqrt(2 * K2 / m)
= sqrt(2 * 91500000 J / 1.5 x 10^4 kg)
≈ sqrt(12200000 m^2/s^2)
≈ 3495 m/s

Therefore, the speed of the airplane after it has traveled 500m is approximately 3495 m/s.

To find the final speed of the airplane after it has traveled 500m, we can use the principle of work and energy.

The work done on an object is equal to the change in its kinetic energy. In this case, the work done on the airplane is equal to the work done by the engine, which is the product of the forward thrust and the displacement of the airplane. So, the work done on the airplane is:

Work = Forward Thrust × Displacement
= (7.5x10^4 N) × (500 m)
= 3.75x10^7 J

The work done on the airplane is equal to the change in its kinetic energy:

ΔKE = Work

Since the airplane is initially moving at 60 m/s, its initial kinetic energy is:

KE_initial = (1/2) × mass × velocity^2
= (1/2) × (1.5x10^4 kg) × (60 m/s)^2
= 5.4x10^6 J

The final kinetic energy of the airplane is:

KE_final = KE_initial + ΔKE
= 5.4x10^6 J + 3.75x10^7 J
= 4.95x10^7 J

Finally, we can determine the final speed of the airplane by equating the final kinetic energy to the kinetic energy formula:

KE_final = (1/2) × mass × velocity_final^2

Rearranging the equation gives:

velocity_final^2 = (2 × KE_final) / mass
= (2 × 4.95x10^7 J) / (1.5x10^4 kg)
= 6.6x10^3

Taking the square root of both sides, we get:

velocity_final = √(6.6x10^3) = 81.2 m/s

Therefore, the final speed of the airplane after traveling 500m is 81.2 m/s.

COMPONENTS OF VECTORS

Vectors are not given all the time in the four directions. For doing calculation more simple sometimes we need to show vectors as in the X, -X and Y, -Y components.

components of vector





For example, look at the vector given below, it is in northeast direction. In the figure, we see the X and Y component of this vector. In other words, addition of Ax and Ay gives us A vector. We benefit from trigonometry at this point. I will give two simple equations which you can use and find the components of any given vector.

components of vector



All vectors can be divided into their components. Now we solve an example and see how we use this technique.

Example Find the resultant vector of A and B given in the graph below. (sin30º=1/2, sin60º=√3/2, sin53º=4/5, cos53º=3/5)

vector example

We use trigonometric equations first and find the components of the vectors then, make addition and subtraction between the vectors sharing same direction.

vector example solution