A 1.7 kg model airplane is flying north at 12.5 m/s initially, and 25 seconds later is observed heading 30 degrees west of north at 25 m/s. What is the magnitude of the average net force on the airplane during this time interval?
A) 0.85 N
B) 0.63 N
C) 1.1 N
D) 1.5 N

Question

Quizplus · Accepted Answer

We can use the equation $\vec{F}_{net} = m\vec{a}$, where $\vec{F}_{net}$ is the net force acting on the airplane, $m$ is its mass, and $\vec{a}$ is its average acceleration during the time interval.

We can break the airplane's initial velocity into its north and west components:

$v_{north} = 12.5 	ext{ m/s}$ 
$v_{west} = 0 	ext{ m/s}$

The final velocity can be broken into its north and west components using trigonometry:

$v_{north} = 25 	ext{ m/s} \cos 30^\circ = 21.7 	ext{ m/s}$ 
$v_{west} = 25 	ext{ m/s} \sin 30^\circ = 12.5 	ext{ m/s}$

The average acceleration in the north direction is:

$a_{north} = \frac{v_{north} - v_{north,0}}{\Delta t} = \frac{21.7 	ext{ m/s} - 12.5 	ext{ m/s}}{25 	ext{ s}} = 0.368 	ext{ m/s}^2$

The average acceleration in the west direction is:

$a_{west} = \frac{v_{west} - v_{west,0}}{\Delta t} = \frac{12.5 	ext{ m/s}}{25 	ext{ s}} = 0.5 	ext{ m/s}^2$

The magnitude of the net force is:

$F_{net} = \sqrt{(ma_{north})^2 + (ma_{west})^2} = \sqrt{(1.7 	ext{ kg})(0.368 	ext{ m/s}^2)^2 + (1.7 	ext{ kg})(0.5 	ext{ m/s}^2)^2} = 1.1 	ext{ N}$

Therefore, the answer is (C).

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