A 15kg model plane flies horizontally at a constant speed of 12.5 m/s and the kinetic energy is 1171J and the potential energy is 3000J what is the new horizontal velocity? Please show work and explain.
To find the new horizontal velocity of the model plane, we need to consider the concept of conservation of mechanical energy. The total mechanical energy of the plane remains constant when there are no external forces acting on it.
The total mechanical energy of an object is given by the sum of its kinetic energy (KE) and potential energy (PE). In this case, the kinetic energy is given as 1171J, and the potential energy is given as 3000J.
Total mechanical energy (E) = KE + PE
Given:
KE = 1171J
PE = 3000J
E = KE + PE
E = 1171J + 3000J
E = 4171J
Since the total mechanical energy remains constant, the new horizontal velocity can be calculated using the formula:
E = (1/2)mv^2
Where:
E = Total mechanical energy
m = Mass of the object
v = Velocity
To rearrange the formula and solve for v, we can multiply both sides of the equation by 2/m:
2E/m = v^2
Taking the square root of both sides:
√(2E/m) = v
Substituting the values:
m = 15kg
E = 4171J
v = √(2(4171J)/(15kg))
v = √(8342J/15kg)
v = √(556.13 m^2/s^2)
v ≈ 23.6 m/s
Therefore, the new horizontal velocity of the model plane is approximately 23.6 m/s.