| Problem statement Summary sheet for work/energy 1 Solution video |
DISCUSSION THREAD
NOTE: This problem is asking for the distance that the rocket travels between when the speed is v1 and when the speed is zero ("at rest").
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Is it better to do the work energy approach or the newton approach as I can see both ways working.
As we have discussed in class, they are fundamentally the same method.
* With Newton, you set up the differential equation, and integrate it over the path.
* With the work/energy equation, the integration of the differential equation has already been done for you.
So, with Newton, you are making an extra effort.
Ok so when we do it with the energy equations we will have to split the forces for the U1->2 integral. Inside those we have the force dot et. Both Ft and Ff are in the same direction of et and so when you take the integral of cos you would have sin. So because they are in the same direction at points and theta equals 0 or 180 that would effectively make those integrals zero right? I think that I missed something as those forces should have an impact.
See Kendall's reply. Also, be sure that you use that F_T and F_f are in OPPOSITE directions (one acts in the direction of motion, and the other opposes the direction of motion).
If you have your axes oriented with the angle, I believe that you can use the property W =Fd for U1->2 similar to example 4.B.2. The only difference is that you have both force from thrust and friction. I don't think that you need to directly do any integration.
Oh ok that's where I was messing up as I was confused with the use of the integrals in that case.
You also have force from the component of gravitational force that is tangent to the path of the rocket, which is just mgsinθ. Other than that yes, this problem can simply be solved with the relation W= Fd very easily since all the motion is in one direction.
The component of gravitational force tangent to the path of the rocket (mgsinθ) isn't relevant here in the approach using energy. This is because it's a conservative force that's already accounted for when doing mgh. When doing W =Fd for U1->2, you only need to take into account the thrust force and friction force.
Since this problem can be solved with both Newton's and Energy method, when we have problems like this one, are we allowed to choose the method that we prefer?
What terms can we leave the answer in? I'm assuming all of the terms on the problem statement (except s and v on the diagram) are fair game.
I left my answer in terms of m,v1,g,Uk,theta,Ft because those were all the givens.
Notice that because the problem specifies "before rest", you cannot assume the final kinetic energy is 0. T2 must factor into your fundamental Work-Energy equation.
This problem is most simply done by shifting your coordinate axis to where the path of the motion is horizontal. Secondly, when determining a final answer, you must leave a v2 in your equation because the problem does specify "before rest". This being said, in your equations, T2 will not be 0. See Aditya's comment.
How are we to solve for v_2 since that is not a variable given in the problem statement?
If you are asking about the potential energy, you will use mgh and plug in the appropriate variables. Keep in mind that the height is the vertical displacement so you will need to be careful when it comes to the sign. If you are referring to the velocity at point 2, I set it to 0 since the problem is asking about it right before it comes to rest. In this instance it is negligible compared to V1 and when subtracting the two, the difference would pretty much be V1.
Just a slight clarification. We are asking for the speed at the instant when the rocket comes to rest, not right before it comes to rest.
Disregard the second part of this comment, TA mistake.
I am seeing comments that say that since it says "before coming to rest", v_2 will not be 0. Should I solve it in terms of v_2 as well or just assume KE_2 = 0.
I solved for the distance traveled when v_2 = 0. I may be wrong, but I think that since the question is asking for the maximum distance traveled before being at rest, that the best way to solve it would be to set v_2 = 0.
Keep in mind that you will also need to use force balance to solve for the Normal force in this problem.
Correct. Assuming you define the x axis as traveling upward along the incline, the summation of forces for the rocket in the y direction will be 0, allowing you to solve for normal force and determine the friction force.
I am seeing people saying T2 does not equal 0. This is NOT true. Read the "NOTE" above the problem in blue font. The note states v2=0 and is "at rest." We can assume the kinetic energy of state 2 to be 0 as well.
I feel like using the work-energy equation is more straightforward than solving out a differential equation when using Newton's method. Is that a fair assessment? Also, can we choose either method or do we have to abide by a specific one in this class?
Dyno: I agree with your assessment. In regard to exams/quizzes/homework, you can get the answer using either work/energy directly or integration of Newton's laws. For "simple problems", you might seen little difference in the complexity of analysis between the two. For problems with even a slight bit of complexity, you may find the effort in integrating Newton to be intractable. I would go with the work/energy equation any time that you are asked to relate a change in speed to a change in position.
I ended up trying both methods, and the work-energy equation was much simpler. I don't think there is a specific way we have to solve the problem, it can be solved either way
I was initially unsure how to write out the kinetics part of the solution since there is an incline, specifically which part I should factor in the angle. Example 4.B.2 from the lecture book is helpful in clearing this up since the system is also on an incline. This homework problem is slightly different since there is a thrust force F_T instead of a spring.
One of the most important things in this question is to carefully keep track of components in both X and Y. Although you might have 4 forces acting on the body, remember to keep track of your non-consecutive ones.
Would it be fair to assume that the forces in the vertical axis cancel each other out where the x axis points along the incline?