KE Work-Energy Theorem Problem question

So, to use your numbers, you should have started with KE_i = 0, and then gone from there, in which case your result will be different. In summary, the conversation discussed two KE Work-Energy Theorem problems related to a golf ball and a fighter jet. The first problem involved calculating the kinetic energy of a golf ball at its highest point and its speed when it is 10.0 m below its highest point. The second problem required finding the work done on a fighter jet by a catapult. The conversation also touched on the concept of energy conservation and the importance of choosing a consistent point for potential energy.
  • #1
djherse
10
0
I am compleatly lost on these two KE Work-Energy Theorem Problem questions any feedback or help would be greatly appreciated. Thank you in advance...

1.)
47.0 g golf ball is driven from the tee with an initial speed of 52.0 m/s and rises to a height of 25.0 m.
(a) Neglect air resistance and determine the kinetic energy of the ball at its highest point.
J
(b) What is its speed when it is 10.0 m below its highest point?
m/s

2.)
A fighter jet is launched from an aircraft carrier with the aid of its own engines and a steam-powered catapult. The thrust of its engines is 1.8 105 N. In being launched from rest it moves through a distance of 87 m and has a kinetic energy of 4.7 107 J at lift-off. What is the work done on the jet by the catapult?


Thanks!
 
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  • #2
djherse said:
I am compleatly lost on these two KE Work-Energy Theorem Problem questions any feedback or help would be greatly appreciated. Thank you in advance...

1.)
47.0 g golf ball is driven from the tee with an initial speed of 52.0 m/s and rises to a height of 25.0 m.
(a) Neglect air resistance and determine the kinetic energy of the ball at its highest point.
J
(b) What is its speed when it is 10.0 m below its highest point?
m/s

2.)
A fighter jet is launched from an aircraft carrier with the aid of its own engines and a steam-powered catapult. The thrust of its engines is 1.8 105 N. In being launched from rest it moves through a distance of 87 m and has a kinetic energy of 4.7 107 J at lift-off. What is the work done on the jet by the catapult?


Thanks!
You should show what you attempted or what you know before any responses will be of help to you.
 
  • #3
well for question 1.) i tried to use KE=.5mv2 and was not right so i tried
.5mv2+mgh also not right i am lost at this point...

And for question 2 i tried to do
4.7x10^7=1.8x10^5+x
also not right any ideas?
 
  • #4
thats a simple view of what i did if you want i can type my numbers and calculations out
 
  • #5
are my equations on the right track or am i just misinformed?
 
  • #6
The point of conservation of energy is that at any point (of your choosing), total energy is conserved. In your case, there's no nonconservative forces so total energy will be a combination of PE and KE. The trick is to find a convenient point to choose
 
  • #7
djherse said:
well for question 1.) i tried to use KE=.5mv2 and was not right so i tried
.5mv2+mgh also not right i am lost at this point...

And for question 2 i tried to do
4.7x10^7=1.8x10^5+x
also not right any ideas?
When there's no friction or applied or other non-cconservative forces acting, energy is conserved. That's the case here, since only gravity acts, and that is a conservative force. Anyway, for the golf ball question, you can now use the conservation of energy principle, Initial KE plus initial PE equals final PE plus final KE. So what's the initial KE? Initial PE? Final PE? Solve for final KE. Use the same approach for part 2 of that question.
 
  • #8
so for the golf ball KE=1/2mV2 + mgh

.5(.047*52^2)+.047*-9.8*25=52.029

Tried it and it worked so thanks guy i was neglecting to make accel due to grav -.

As far as the jet problem i am still lost...
 
  • #9
ok having problems with part b of golf ball problem

47.0 g golf ball is driven from the tee with an initial speed of 52.0 m/s and rises to a height of 25.0 m.
(a) Neglect air resistance and determine the kinetic energy of the ball at its highest point.
J
(b) What is its speed when it is 10.0 m below its highest point?
m/s

I have used
KEi+PEi = KEf+PEf so
52.029= .0235*x^2 + -6.909

solve for x you get x= 50.07993

However this is not right what am i doing wrong?
 
  • #10
bump...
 
  • #11
bump...
 
  • #12
djherse said:
ok having problems with part b of golf ball problem

47.0 g golf ball is driven from the tee with an initial speed of 52.0 m/s and rises to a height of 25.0 m.
(a) Neglect air resistance and determine the kinetic energy of the ball at its highest point.
J
(b) What is its speed when it is 10.0 m below its highest point?
m/s

I have used
KEi+PEi = KEf+PEf so
52.029= .0235*x^2 + -6.909

solve for x you get x= 50.07993

However this is not right what am i doing wrong?
If you used the KE at the top as your initial point, and PE initial equal to 0 at that point, then when you look at the final PE, you must use h = - 10, not -15 as you used. Alternatively, you could have started at the very beginning, where KE_i =1/2(.047)(52)^2, PE_i = 0, and PE_f = + .047(9.8)(15), which will yield the same result when you use the equation. You can choose any point as PE=O, but then you must be consistent thereafter.
 
Last edited:

Related to KE Work-Energy Theorem Problem question

1. What is the KE Work-Energy Theorem?

The KE Work-Energy Theorem is a fundamental principle in physics that states that the kinetic energy of an object is equal to the work done on the object. This means that the amount of energy an object has due to its motion is equal to the amount of work needed to change its motion.

2. How is the KE Work-Energy Theorem calculated?

The KE Work-Energy Theorem is calculated by multiplying the mass of an object by its velocity squared, divided by two. This can be represented by the equation KE = 1/2mv^2, where KE is kinetic energy, m is mass, and v is velocity.

3. What units are used to measure KE and work?

Kinetic energy is typically measured in joules (J), while work is also measured in joules (J). However, in some cases, other units such as foot-pounds (ft-lb) or calories (cal) may be used to measure these quantities.

4. How is the KE Work-Energy Theorem applied in real-world problems?

The KE Work-Energy Theorem can be applied in a variety of real-world problems, such as calculating the speed of a rollercoaster at different points on its track, determining the energy needed to launch a rocket into space, or calculating the force needed to stop a moving car. It is a useful tool for understanding and predicting the motion and energy of objects in our everyday lives.

5. What are some common misconceptions about the KE Work-Energy Theorem?

One common misconception about the KE Work-Energy Theorem is that it only applies to objects with constant velocity. In reality, the theorem applies to any object with a changing velocity, as long as the net work done on the object is known.

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