A model helicopter and its load have a combined mass of 3.5 kg. Find the upward average force, in Newtons, exerted by the air if the helicopter is originally moving downward at 10 m/s and is brought to rest with constant acceleration in a distance of 50 m.
Fnet = ma
F-mg = ma
a = constant = 0
F = mg
40 N
Well, well, well! Looks like the model helicopter is in quite a sticky situation!
To find the upward average force exerted by the air, we first need to calculate the deceleration the helicopter experienced.
Using the equation:
v^2 - u^2 = 2as,
where v is the final velocity (0 m/s in this case), u is the initial velocity (-10 m/s downward), a is the acceleration, and s is the distance (50 m), we can rearrange the equation to solve for a:
a = (v^2 - u^2) / 2s.
Substituting in the known values:
a = (0^2 - (-10)^2) / (2 * 50).
Simplifying this equation:
a = (0 - 100) / 100.
a = -1 m/s^2.
So, the helicopter experienced an acceleration of -1 m/s^2, indicating that it was decelerating at a rate of 1 m/s^2.
Now, to calculate the upward average force exerted by the air, we can use Newton's second law:
F = ma,
where F is the force, m is the mass, and a is the acceleration.
Substituting the given values:
F = (3.5 kg) * (-1 m/s^2).
F = -3.5 N.
Wait a minute, did I read that right? Negative force? That's quite odd!
But fear not, my friend! The negative sign indicates that the force is acting in the opposite direction of motion, which in this case is up! So, the average upward force exerted by the air is 3.5 Newtons.
Now, let's hope the helicopter gets back to its normal high-flying shenanigans soon!
To find the upward average force exerted by the air on the model helicopter, we can use the equation for acceleration:
v^2 = u^2 + 2as
where,
v = final velocity (0 m/s since the helicopter is brought to rest)
u = initial velocity (-10 m/s, negative because it is moving downward)
a = acceleration
s = distance (50 m)
Rearranging the equation, we have:
a = (v^2 - u^2) / (2s)
= (0 - (-10)^2) / (2 * 50)
= (-100) / 100
= -1 m/s^2
The negative sign indicates that the acceleration is in the opposite direction of the initial velocity, which means it is an upward acceleration.
Now, we can calculate the upward average force exerted by the air using Newton's second law:
F = ma
where,
F = force
m = mass (combined mass of the helicopter and its load = 3.5 kg)
a = acceleration (-1 m/s^2)
Plugging in the values:
F = 3.5 kg * (-1 m/s^2)
= -3.5 N
The negative sign indicates that the force is upward, opposite to the direction of the helicopter's motion. Therefore, the upward average force exerted by the air is 3.5 Newtons.
To solve this problem, we need to apply Newton's second law of motion: F = ma, where F is the net force, m is the mass, and a is the acceleration.
However, in this case, we need to find the average force. The formula for average force is given by the equation: F_avg = Δp / Δt, where F_avg is the average force, Δp is the change in momentum, and Δt is the change in time.
Since the helicopter is brought to rest with a constant acceleration, we can use the equation of motion: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
Given:
Initial velocity, u = -10 m/s (negative sign indicating downward motion)
Final velocity, v = 0 m/s (helicopter brought to rest)
Displacement, s = 50 m
Using the equation of motion, we can find the acceleration:
v^2 = u^2 + 2as
0^2 = (-10)^2 + 2a(50)
0 = 100 + 100a
100a = -100
a = -1 m/s^2 (negative sign indicating upward acceleration)
Now, we can calculate the change in momentum using the formula Δp = mΔv:
Δp = m(v - u) [since u is negative]
Δp = 3.5 kg * (0 - (-10) m/s)
Δp = 3.5 kg * 10 m/s
Δp = 35 kg·m/s
Since Δt is not given, let's assume it takes 5 seconds to bring the helicopter to rest at a constant acceleration.
Now, we can calculate the average force:
F_avg = Δp / Δt
F_avg = 35 kg·m/s / 5 s
F_avg = 7 N
Therefore, the average upward force exerted by the air on the helicopter is 7 Newtons.