Francesca, who likes physics experiments, dangles her watch from a thin piece of string while the jetliner she is in takes off from JFK Airport. She notices that the string makes an angle of 25° with respect to the vertical as the aircraft accelerates for takeoff, which takes about 18 s. Estimate the takeoff speed of the aircraft.
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What follows is a calculated question with 1 blanks. A box is given a push so that it slides across the floor. Given that the coefficient of kinetic friction is 0.4 and the push imparts an initial speed of 2.6 m/s, the total distance travelled by the box is m.
To estimate the takeoff speed of the aircraft, we can use trigonometry and the physics concept of centripetal acceleration.
First, let's analyze the situation. Francesca is in an aircraft at JFK Airport, and as the plane accelerates for takeoff, she notices that her watch, which is dangled from a string, makes an angle of 25° with respect to the vertical. This angle can be assumed to be the angle between the string and the vertical axis.
Now, let's break down the forces at play. The force of gravity acts vertically downward, while the tension in the string acts radially inward, towards the center of the circular motion.
Using trigonometry, we can relate the tension (T) in the string to the component of gravitational force perpendicular to the string. The component of the gravitational force perpendicular to the string is given by:
F_perpendicular = m * g * sin(θ)
Where:
m is the mass of the watch,
g is the acceleration due to gravity (approximately 9.8 m/s^2),
θ is the angle the string makes with the vertical axis (25°).
Since the watch is in equilibrium, the tension in the string (T) should be equal to the force perpendicular to the string (F_perpendicular). Therefore, we can write:
T = m * g * sin(θ)
Now, we can relate the tension in the string to the centripetal force acting on the watch. The centripetal force is given by:
F_centripetal = m * a
Where:
m is the mass of the watch,
a is the centripetal acceleration of the watch.
The centripetal acceleration can be related to the angular velocity (ω) and the radius of the circular motion (r) by the formula:
a = ω^2 * r
Since the string is assumed to be thin, and hence effectively weightless, the tension in the string is providing the required centripetal force for the circular motion. Therefore, we can equate the tension in the string (T) to the centripetal force (F_centripetal):
T = F_centripetal
Substituting the expressions for T and F_centripetal, we get:
m * g * sin(θ) = m * ω^2 * r
In this case, the angle θ is 25°, and the radius (r) is equal to the length of the string.
Now, we need to find the angular velocity (ω). To do this, we can use the relationship between angular velocity, linear velocity (v), and radius (r):
ω = v / r
Rearranging the equation, we get:
v = ω * r
Substituting this into our previous equation, we get:
m * g * sin(θ) = m * (v^2 / r) * r
Simplifying the equation further:
g * sin(θ) = v^2
Finally, we can solve for the takeoff speed (v). Since the aircraft is accelerating for takeoff, we assume that the speed is changing. Therefore, we need to consider the average takeoff speed during the 18-second interval:
v_average = ∆d / ∆t
Where:
∆d is the change in distance (which is equal to r), and
∆t is the time interval (which is equal to 18 s).
Rearranging the equation, we get:
v_average = r / ∆t
Substituting the expression for v_average into our previous equation, we get:
g * sin(θ) = (r / ∆t)^2
Now, we can plug in the known values:
g ≈ 9.8 m/s^2
θ = 25°
∆t = 18 s
Solving for r, we need to find the length of the string.
Unfortunately, the length of the string is not given in the question. Without the length of the string, it is not possible to estimate the takeoff speed of the aircraft accurately.
To estimate the takeoff speed, we need the length of the string. But if we assume the length of the string is known, we can substitute the values into the equation and solve for v, giving us an estimate of the takeoff speed.