An airplane of mass 1.6 x 10^4 kg is moving at 58 m/s. The pilot then revs up the engine so that the forward thrust by the air around the propeller becomes 7.0 x 10^4 N. If the force exerted by air resistance on the body of the airplane has a magnitude of 6.0 x 10^4 N, find the speed of the airplane after it has traveled 500 m. Assume that the airplane is in level flight throughout this motion.
77 m/s
To solve this problem, we can use the principle of conservation of energy. The initial kinetic energy of the airplane is equal to the work done by the forward thrust of the propeller minus the work done against air resistance. Let's break it down step-by-step:
Step 1: Calculate the initial kinetic energy of the airplane.
Kinetic energy (KE) is given by the formula KE = (1/2)mv^2, where m is the mass of the airplane and v is its velocity.
Plugging in the given values: mass (m) = 1.6 x 10^4 kg, velocity (v) = 58 m/s.
KE = (1/2)(1.6 x 10^4 kg)(58 m/s)^2
Step 2: Calculate the work done by the forward thrust of the propeller.
Work (W) is given by the formula W = Fd, where F is the force and d is the displacement.
Plugging in the given values: force (F) = 7.0 x 10^4 N, displacement (d) = 500 m.
W = (7.0 x 10^4 N)(500 m)
Step 3: Calculate the work done against air resistance.
Plugging in the given value: force (F) = 6.0 x 10^4 N, displacement (d) = 500 m.
W = (6.0 x 10^4 N)(500 m)
Step 4: Apply the principle of conservation of energy.
According to the principle of conservation of energy, the initial kinetic energy is equal to the work done by the forward thrust minus the work done against air resistance.
KE = W_forward - W_air_resistance
Step 5: Calculate the final velocity of the airplane.
Rearrange the equation from step 4 to find the final velocity (v):
KE = W_forward - W_air_resistance
(1/2)(1.6 x 10^4 kg)(v^2) = (7.0 x 10^4 N)(500 m) - (6.0 x 10^4 N)(500 m)
Solve for v.
By following these steps, you can calculate the speed of the airplane after it has traveled 500 m.
To find the final speed of the airplane after it has traveled 500 m, we can use the concept of work-energy theorem.
The work-energy theorem states that 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 given by the net force acting on it multiplied by the distance traveled. The change in kinetic energy of the airplane will be equal to the work done on it.
The net force acting on the airplane is the difference between the forward thrust by the air around the propeller and the force of air resistance. So, the net force is (7.0 x 10^4 N - 6.0 x 10^4 N) = 1.0 x 10^4 N.
The work done on the airplane is the net force multiplied by the distance traveled. So, the work done is (1.0 x 10^4 N) * (500 m) = 5.0 x 10^6 joules.
According to the work-energy theorem, the work done is equal to the change in kinetic energy. So, the change in kinetic energy is 5.0 x 10^6 joules.
The initial kinetic energy of the airplane is given by (1/2) * mass * (initial velocity)^2. So, the initial kinetic energy is (1/2) * (1.6 x 10^4 kg) * (58 m/s)^2 = 3.7584 x 10^6 joules.
The final kinetic energy of the airplane is equal to the sum of the initial kinetic energy and the change in kinetic energy. So, the final kinetic energy is 3.7584 x 10^6 joules + 5.0 x 10^6 joules = 8.7584 x 10^6 joules.
The final speed of the airplane can be found using the equation for kinetic energy: (1/2) * mass * (final velocity)^2 = final kinetic energy.
Rearranging the equation, we get (final velocity)^2 = (2 * final kinetic energy) / mass.
Plugging in the values, we have (final velocity)^2 = (2 * 8.7584 x 10^6 joules) / (1.6 x 10^4 kg).
Simplifying, we find (final velocity)^2 ≈ 10932 m^2/s^2.
Taking the square root of both sides of the equation, we get the final velocity ≈ 104.6 m/s (rounded to the nearest tenth).
Therefore, the speed of the airplane after it has traveled 500 m is approximately 104.6 m/s.