Which Solution is Incorrect and Why?

  • Thread starter Thread starter misterpickle
  • Start date Start date
AI Thread Summary
The discussion centers on a discrepancy between a solutions manual and a student's submission regarding the work done by a particle. Both solutions appear mathematically correct, but they yield different results. The key issue is that the student's approach uses the work formula for constant forces, which is not applicable in this scenario. The correct method involves integrating power over time, confirming that the student's solution is indeed incorrect. Ultimately, the discussion concludes that the student's method fails due to its reliance on a constant force assumption.
misterpickle
Messages
9
Reaction score
0
I'm a TA and there is an inconsistency in what the solutions manual states and what the students are turning in, but both seem correct.

The Latex code isn't working properly for the post, but it is easy to see what equations are used

Homework Statement


A particle of mass m moves along the x axis. Its position varies with time according to $x$=2t^3 - 4t^2 , where x is in meters and t is in seconds. Find

a. The velocity and acceleration of the particle as functions of t.

b. The power delivered to the particle as a function of t

c. the work done by the net force from t=0 $to$ t=t_{1}

Homework Equations


The first 2 parts are easy and the answers are given by
$v$=6t^2 - 8t $;$ $a$=12t - 8 $;$ $P$=power=8mt(9t^2 - 18t + 8)

The Attempt at a Solution



The problem is that both solutions are correct mathematically, but they are different. Therefor one must be wrong and I need someone to tell me which on, and WHY?

Solution manual
$w$=\int_0^t \! P \, dt<br /> $w$=\int_0^t \! 8m(9t^3 - 18t^2 + 8t) \, dt<br /> $w$=2mt^{2}(9t^2-24t+16)
Student Solution
$w$=F\Delta x=ma\Delta x = m(12t - 8)(2t^3 - 4t^2)
This becomes,
$w$=m(24t^4 - 64t^3 + 32t)
Which one of these is wrong, since the math is right on both of them and the equations are valid for power?
 
Physics news on Phys.org
The second equation (Work = force*distance moved) is only valid for constant forces, in fact you can derive it from the first equation by letting P = F*dx/dt for some constant F.
 
So the student solution is incorrect?
 
misterpickle said:
Student Solution
$w$=F\Delta x=ma\Delta x = m(12t - 8)(2t^3 - 4t^2)

misterpickle said:
So the student solution is incorrect?
Yes, it is incorrect. The correct equation is
W = F dx

Which, as marmoset said, equals F Δx only for a constant force.
 
misterpickle said:
So the student solution is incorrect?

students looks wrong to me
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Back
Top