Calculating Aeroplane Speed and Archer's Bow Work: Physics Problems Solved

  • Thread starter DNA
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In summary, the aeroplane of mass 1.5 x 104 kg initially moves at 60ms-1. After the pilot revs up the engine, the forward thrust becomes 7.5 x 104 N and the air resistance is 4.0 x 104 N. Assuming level flight, the final speed of the aeroplane after traveling 500m can be found using the initial velocity, net force, mass, and kinematics formulas. The equivalent spring constant and work done in pulling back an archer's bow can also be calculated using the force equation and work formula.
  • #1
DNA
3
0
Can someone please help me with the following?

An aeroplane of mass 1.5 x 104 kg is moving at 60ms-1.
The pilot then revs up the engine so that the forward thrust of the propeller becomes 7.5 x 104 N. If the force of the air resistance has a magnitude
of 4.0 x 104 N, find the speed of the aeroplane after it has traveled 500m. Assume that the aeroplane is in level flight throughout this motion.

An archer pulls her bow string back 0.400m by exerting a force that increases uniformly from zero to 230N.
a) What is the equivalent spring constant of the bow?
b) How much work is done in pulling the bow?
 
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  • #2
What have you tried to do to solve these problems?
 
Last edited:
  • #3
The Archer problem

To solve part a.)

k= mg / d So d = 4.000 meters and m= 230 Netwons

Note that 1 kg = 9.8 Newtons. Therfore, 226 N = 23.469 kg .

Now plug in and solve for k = (( 23.469 kg ) * ( 9.8 m/s^(2)))/ ( 4.000 m ) = 57.5 N/m
 
  • #4
Correction Therefore 230 N = 23.469 kg
 
  • #5
ccwitt said:
The Archer problem

To solve part a.)

k= mg / d So d = 4.000 meters and m= 230 Netwons

Note that 1 kg = 9.8 Newtons. Therfore, 226 N = 23.469 kg .

Now plug in and solve for k = (( 23.469 kg ) * ( 9.8 m/s^(2)))/ ( 4.000 m ) = 57.5 N/m

What are you doing?

[tex] PE_{spring} = \frac{1}{2}kx^2 = W_{spring} [/tex]
[tex] F_{spring} = kx [/tex]

to answer part a, just sub into the force equation and solve for k. To answer part b, use Work formula I wrote. As for the first question, just subtract your thrust force from your air resistance force, and then you have your net force. You have mass, so you can slove for acceleration. Once you have this, use the kinematics formulas to solve for final speed.

Regards,

Nenad
 
  • #6
Nenad said:
What are you doing?

[tex] PE_{spring} = \frac{1}{2}kx^2 = W_{spring} [/tex]
[tex] F_{spring} = kx [/tex]

to answer part a, just sub into the force equation and solve for k. To answer part b, use Work formula I wrote. As for the first question, just subtract your thrust force from your air resistance force, and then you have your net force. You have mass, so you can slove for acceleration. Once you have this, use the kinematics formulas to solve for final speed.

Regards,

Nenad

In part a.) it works the same a a spring vertically hung undergoing a load "mg" with a distance d.

In the case of the Archer we just apply Hooke's Law like in a spring to give
|F (sub s) | = k*d = mg this implies k = mg/d

So in part a.) Its really k = 230 N / 0.400m =575 N/m This is a clerical error made by myself because I was in a hurry. I should have been more carefull in writing the response.
 
Last edited:
  • #7
and as for the aeroplane...

DNA said:
Can someone please help me with the following?

An aeroplane of mass 1.5 x 104 kg is moving at 60ms-1.
The pilot then revs up the engine so that the forward thrust of the propeller becomes 7.5 x 104 N. If the force of the air resistance has a magnitude
of 4.0 x 104 N, find the speed of the aeroplane after it has traveled 500m. Assume that the aeroplane is in level flight throughout this motion.

you have the initial velocity of the plane, you have the new tractive force of the engine/propeller minus the air resistive force of the plane, you have f=mA, a=dv/dt, and some other things to get you to from the initial speed to the final speed, how long it took to go 500m and you can assume all other things you want, including ignoring gravity...
:smile:
 

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