Practice Problem 0175
Another plane is flying East and maintains a constant altitude. The lift force acting on the plane is
100000 N. The initial velocity of the plane is 100 m/s [E], when it begins to accelerate at a constant
rate. If the force due to drag acting on the plane is 40 000 N, and the applied force of thrust acting on
the plane is 52 000N, find the time it takes for the plane to achieve a velocity of 115 m/s [E].
This situation can be thought of as two independent problems. In the vertical direction (y direction),
the forces due to lift and gravity are acting. In the horizontal direction (x direction), drag and thrust are
the forces that are acting.
FLIFT = 100,000 N
FThrust = 52,000 N
∆t = ???
V1 = 100 m/s [E]
ay = 0
V2 = 115 m/s [E]
FDrag = 40,000 N
We know that in the y direction (vertical direction), acceleration is 0 since the plane maintains a
constant altitude.
Therefore, to solve for time we need to begin with the x direction (horizontal direction) and try to solve
for acceleration.
FNETX = FThrust – FDrag
ma = 52,000- 40,000
ma = 12,000
The problem is that we have one equation with two unknowns. Turning our attention to the vertical
direction (y direction)
FNETY = FLIFT - FG
0 = 100,000 – (mg)
mg = 100,000
(9.8)g = 100,000
m = 10,204.08163 kg
Now that we know the mass, we will return to our equation involving the horizontal direction to
determine the acceleration.
FNETX = FThrust – FDrag
ma = 12,000
10,204.08163a = 12,000
a = 1.176 m/s2
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Now using our kinematic equations to solve for time,
a = V2 – V1
∆t
∆t = V2 – V1
a
∆t = 115 – 110
1.176
∆t = 12.76’s