Question

8) When an F-14 airplane takes off an aircraft carrier it is literally catapulted of the flight deck.
The plane's final speed at take-off is 68.2 m/s. The F-14 starts from rest. The plane accelerates
in 2 seconds and has a mass of 29 545 kg. What is the total force that gets the F-14 in the air? 9)A sports car accelerates from 0 to 27 m/s, 60 mph, in 6.3 seconds. The car exerts a force of
4106 N. What is the mass of the car?
10)A boy can accelerate at 1.00 m/s? over a short distance. If the boy were to take an energy
pill and suddenly have the ability to accelerate at 5.6 m/s“, then how would his new energy-pill-
force compare to his earlier force? If the boy's earlier force was 45 N, what is the size of his
energy-pill-force?
12) A cartoon plane with four engines can accelerate at 8.9 m/s- when one engine is running.
What is the acceleration of the plane if all four engines are running and each produces the same
force?

Answer 1

8) $1.0\cdot 10^6 N$

9) 958 kg

10) $F_2=5.6 F_1, F_2=252 N$

11) $35.6 m/s^2$

Explanation:

8)

The total force exerted on an object is equal to the rate of change of momentum of the object; mathematically:

$F=\frac{\Delta p}{\Delta t}$

where

F is the force

$\Delta p$ is the change in momentum

$\Delta t$ is the time elapsed

The change in momentum can be written as

$\Delta p = m(v-u)$

where

m is the mass

v is the final velocity

u is the initial velocity

For the plane in this problem:

u = 0 (it starts from rest)

v = 68.2 m/s

$\Delta t = 2 s$

m = 29545 kg

Therefore, the force exerted on the F-14 is:

$F=\frac{(29545)(68.2-0)}{2}=1.0\cdot 10^6 N$

9)

As in the previous problem, the force exerted on the car can be written as

$F=\frac{\Delta p}{\Delta t}=\frac{m(v-u)}{\Delta t}$

where in this problem, we have:

u = 0 is the initial velocity of the car

v = 27 m/s is the final velocity of the car

$\Delta t = 6.3 s$ is the time elapsed during the acceleration of the car

F = 4106 N is the force exerted on the car

Re-arranging the equation for m, we can find the mass of the car:

$m=\frac{F\Delta t}{v-u}=\frac{(4106)(6.3)}{27-0}=958 kg$

So, the mass of the  car is 958 kg.

10)

The relationship between force exerted on a body and acceleration of the body is given by Newton's second law of motion:

$F=ma$

where

F is the force

m is the mass

a is the acceleration

At the beginning, with a certain force $F_1$, the acceleration of the boy is

$a_1=1.00 m/s^2$

So we have

$F_1=ma_1$ (1)

Later, after the boy has taken the pill, the force is $F_2$, while the acceleration is now

$a_2=5.6 m/s^2$

So we have

$F_2=ma_2$ (2)

To do a comparison between the two forces, we divide (2) by (1), and we get:

$\frac{F_2}{F_1}=\frac{a_2}{a_1}=\frac{5.6}{1}=5.6$

So,

$F_2=5.6 F_1$

Which means that the force in the 2nd case is 5.6 times larger than the force in the 1st case.

We are also told that the boy's earlier force is

$F_1=45 N$

Therefore, the force in the 2nd case is:

$F_2=5.6(45)=252 N$

12)

For this problem we can use again Newton's second law of motion:

$F=ma$

where

F is the force

m is the mass

a is the acceleration

At the beginning, with one engine only running, the acceleration of the plane is

$a=8.9 m/s^2$

And we can write it as

$a=\frac{F}{m}$

where F is the force exerted by 1 engine only.

Later, when  all 4 engines are running, the new force will be 4 times:

F' = 4F

Therefore, the new acceleration will be:

$a'=\frac{F'}{m}=\frac{4F}{m}=4(\frac{F}{m})=4a$

So, 4 times the previous acceleration; and therefore:

$a'=4a=4(8.9)=35.6 m/s^2$