A 15 kg model airplane flies horizontally at point A that is 20 m about point B. It has a speed of 19.993 m/s at point A. It then flies horizontally at point B. If the loss of potential energy is reflected as an increase in kinetic energy, what is the speed of the model airplane at point B?
To solve this problem, we need to apply the principle of conservation of energy, which states that the total mechanical energy of a system remains constant when only conservative forces (like gravity) are acting on it.
In this case, the loss of potential energy of the model airplane (due to its change in height) is reflected as an increase in its kinetic energy. The potential energy at point A is given by the formula:
PE(A) = m * g * h(A)
where m is the mass of the model airplane, g is the acceleration due to gravity, and h(A) is the height at point A.
Since the airplane is flying horizontally, there is no change in its height between points A and B. Therefore, the potential energy at point B is the same as at point A:
PE(B) = PE(A) = m * g * h(A)
Now, let's find the kinetic energy at point A. Kinetic energy is given by the formula:
KE(A) = (1/2) * m * v(A)^2
where m is the mass of the model airplane and v(A) is its speed at point A.
Given the values in the problem statement, we can calculate the potential energy and kinetic energy at point A.
PE(A) = m * g * h(A) = 15 kg * 9.8 m/s^2 * 20 m = 2940 J
KE(A) = (1/2) * m * v(A)^2 = (1/2) * 15 kg * (19.993 m/s)^2 = 2998.5 J
Since the loss of potential energy is reflected as an increase in kinetic energy, we can equate the potential energy at point B to the sum of the kinetic energies at points A and B:
PE(B) = KE(A) + KE(B)
m * g * h(B) = (1/2) * m * v(A)^2 + (1/2) * m * v(B)^2
Canceling out the mass (m) from both sides of the equation:
g * h(B) = (1/2) * v(A)^2 + (1/2) * v(B)^2
Rearranging the equation to solve for the speed of the model airplane at point B (v(B)):
v(B)^2 = 2 * (g * h(B) - (1/2) * v(A)^2)
v(B) = sqrt(2 * (g * h(B) - (1/2) * v(A)^2))
Now, substitute the given values to find the speed of the model airplane at point B. The height at point B is the same as at point A, and g is the acceleration due to gravity (9.8 m/s^2).
v(B) = sqrt(2 * (9.8 m/s^2 * 20 m - (1/2) * (19.993 m/s)^2))
v(B) = sqrt(2 * (196 m^2/s^2 - 199.924 m^2/s^2))
v(B) = sqrt(2 * (-3.924 m^2/s^2))
Since we cannot take the square root of a negative value within the realm of real numbers, it appears that there may be an error in the problem statement or the calculations provided.