Engineering Physics (Work done and energy) Help

In summary, the conversation discusses questions 2 and 5 regarding the work done by a catapult and the kinetic energy of a particle in circular motion. For question 2, it is determined that the engine and catapult both do work on the aircraft, with the engine doing 23000 x 87 of work and the total work being 4.5 x 10^7 J. Therefore, the work done by the catapult must be 4.5 x 10^7 - 23000 x 87 = 42999000. For question 5, the speed of the particle is unknown and the result of the calculation for question 2 is surprising as the catapult does significantly more work than the engine.
  • #1
Benny T
4
0
Q2 AND Q5 are the questions I struck in at the minute. For Q 2 , I wonder i could simply use change of kinetic energy=Work done, does the catapult gain the same amount of energy as the Jet fighter?? For Q.5, i have no idea where to start, could someone at least give me some kind of direction??

Question 2:2. Jet fighter J-15 is launched from an aircraft carrier with the aid of its own engines and a catapult. J-15 is 6020 kg and the thrust of the engines is 23000N. At lift off, J-15 has a kinetic energy of 4.5x107 J and has moved 87m from the rest. What is the work done by the catapult?

Question 5:5. 1 A small particle (so the volume can be ignored) with mass m is having a circular motion with constant speed v and radius R. Determine it’s kinetic energy. 5. 2 A thin, solid, uniform disk of radius R and mass M is rotating constantly around the vertical symmetric axis as shown below. The constant speed at the rim is v. Calculate the kinetic energy of the disc. (Hint: A rotating disc is composed of particles as described in 5.1 with various R and v(R))
 

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  • #2
Benny T said:
, does the catapult gain the same amount of energy as the Jet fighter??
The catapult and engine both do work on the aircraft. How much work is done by the engine? How much total work is done on the aircraft?

For Q5, how fast is the particle travelling?
 
  • #3
Q.2 engine has done 23000x87 of work , total work has to be 4.5 x10^7 J as this is the lift off energy(NOT SURE)? so work done by catapult should be 4.5x10^7-23000x87=42999000
 
  • #4
thank you for the reply!
 
  • #5
Benny T said:
Q.2 engine has done 23000x87 of work , total work has to be 4.5 x10^7 J as this is the lift off energy(NOT SURE)? so work done by catapult should be 4.5x10^7-23000x87=42999000
Looks right...
But too many significant figures compared with the given data.
 
  • #6
The result is surprising as the catapult actually do much more work than the engine itself (almost 21 times more)!
 

1. What is work done in engineering physics?

Work done in engineering physics is the transfer of energy from one system to another. It is the product of the force applied to an object and the distance it moves in the direction of the force. It is usually measured in joules (J).

2. How is work related to energy in engineering physics?

Work and energy are closely related in engineering physics. Work is the process of transferring energy from one form to another. Energy is the ability to do work. This means that the more work that is done, the more energy is transferred.

3. What are the different types of energy in engineering physics?

The main types of energy in engineering physics are kinetic energy, potential energy, and thermal energy. Kinetic energy is the energy of motion, potential energy is stored energy, and thermal energy is the energy of heat.

4. How is power related to work in engineering physics?

Power is the rate at which work is done. It is measured in watts (W) and is the amount of work done in a given amount of time. The more powerful a system is, the faster it can do work.

5. What are some real-life examples of work done in engineering physics?

Some examples of work done in engineering physics include lifting weights, riding a bike, heating a room, and charging a phone. In all of these cases, energy is being transferred from one system to another through the process of work.

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