A plane weighing 236 kN (24.1 tons) lands on an aircraft carrier. The plane is moving horizontally at 66.8 m/s (149.4 mi/h) when its tailhook grabs hold of the arresting cables. The cables bring the plane to a stop in a distance of 84.0 m.
What is the force (assumed constant) exerted on the plane by the cables? Enter a negative value if the force exerted on the plane is opposite to its direction of motion. Enter a positive value if the force exerted on the plane is in the direction of motion.
To solve this problem, we can use the work-energy principle, which states that the net work done on an object is equal to its change in kinetic energy. In this case, the net work done on the plane is the force exerted by the cables multiplied by the distance over which the force acts, and the change in kinetic energy is equal to the initial kinetic energy (due to the plane's motion) minus the final kinetic energy (when it comes to a stop). We can set these two expressions equal to each other and solve for the force:
Force x distance = (1/2) x mass x (final velocity)^2 - (1/2) x mass x (initial velocity)^2
where mass = 236 kN / 9.81 m/s^2 = 24.06 metric tons
Plugging in the given values, we get:
Force x 84.0 m = (1/2) x 24.06 t x (0 m/s)^2 - (1/2) x 24.06 t x (66.8 m/s)^2
Simplifying and solving for the force, we get:
Force = - 572.7 kN
The negative sign indicates that the force exerted on the plane is opposite to its direction of motion. Therefore, we can conclude that the arresting cables exert a force of 572.7 kN in the opposite direction to the plane's motion to bring it to a stop.
Plane mass = (236,000 Newtons / 9.81 m/s^2) = 24,057 kg
So initial momentum =
24,057 * 66.8 meters/second
= 1,607,000 kilograms meters/second
time to stop = 84.0 meters / 33.4 meters/sec = 2.515 seconds
av force = change in momentum / time =1,607,000 /2.515 =639,000 N
Your approach is correct, but there is an error in your calculation of the time. The speed of the plane is given in miles per hour (mph), not meters per second (m/s), so we need to convert it first:
66.8 m/s = (66.8 / 2.237) mph ≈ 149.4 mph
Using this value, we can calculate the time to stop as:
time = distance / speed = 84.0 m / (149.4 mph * 0.44704 m/s/mph) ≈ 1.48 s
Now we can use the impulse-momentum theorem to calculate the average force exerted on the plane by the arresting cables:
average force = change in momentum / time
The initial momentum of the plane is:
p = m * v = 24,057 kg * 66.8 m/s ≈ 1,607,000 kg m/s
The final momentum is zero, since the plane comes to a stop. Therefore, the change in momentum is equal to the initial momentum.
Substituting the values, we get:
average force = 1,607,000 kg m/s / 1.48 s ≈ 1,084,000 N
The positive sign indicates that the force exerted on the plane is in the direction of motion (opposite to the direction of the cable's force). Therefore, we can conclude that the arresting cables exert an average force of approximately 1,084 kN in the opposite direction to the plane's motion to bring it to a stop.
but it said:
" horizontally at 66.8 m/s (149.4 mi/h) "
so the average speed during the stop is 33.4 m/s
The force is negative of course
You are correct, I apologize for the mistake in my previous response. The speed given in the problem is indeed the initial speed, and we need to divide it by 2 to get the average speed during the stop (assuming constant acceleration). Therefore, the average speed during the stop is:
v = 66.8 m/s / 2 = 33.4 m/s
Using this value, we can calculate the time to stop as:
time = distance / speed = 84.0 m / 33.4 m/s ≈ 2.51 s
Now we can use the impulse-momentum theorem to calculate the average force exerted on the plane by the arresting cables:
average force = change in momentum / time
The initial momentum of the plane is:
p = m * v = 24,057 kg * 66.8 m/s ≈ 1,607,000 kg m/s
The final momentum is zero, since the plane comes to a stop. Therefore, the change in momentum is equal to the initial momentum.
Substituting the values, we get:
average force = 1,607,000 kg m/s / 2.51 s ≈ -640,000 N
The negative sign indicates that the force exerted on the plane is opposite to its direction of motion. Therefore, we can conclude that the arresting cables exert an average force of approximately 640 kN in the opposite direction to the plane's motion to bring it to a stop. Thank you for pointing out my mistake.