Calculating K in the XYZ Space: A Homework Challenge

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Homework Help Overview

The problem involves a particle of mass 1 kg projected in XYZ space under the influence of gravity. The initial velocity and components of acceleration are provided, with the goal of determining the value of K based on the total work done over a specified time interval.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to show attempts at the solution to receive help, with some suggesting the citation of relevant equations. One participant provides a detailed attempt at calculating the work done, while others point out potential errors in calculations and seek alternative methods to avoid mistakes.

Discussion Status

Some participants have offered corrections to calculations and shared insights on using mathematical software to minimize errors. There is an ongoing exploration of different methods to approach the problem, but no consensus has been reached regarding the correctness of the proposed solutions.

Contextual Notes

Participants note that the format of the homework requires K to be an integer between 0 and 9, and there is an emphasis on showing work to facilitate assistance.

particlemania
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Homework Statement


A particle of mass 1kg is projected in XYZ space, where Gravity (g=10m/s2) acts in -\hat{k} direction. The initial velocity of a particle is \vec{u}=(-3\hat{i}+4\hat{j})m/s.
x-component of acceleration = 3t/4
y-component of acceleration = -1 - 3t/4
If total work done in interval t=0 to t=4 seconds is 90K Joules, then find the value of K.

[The format of answering requires K to be an integer between 0 and 9 (inclusive)]


Homework Equations


Basic Kinematic definitions with complicated level of Calculus


The Attempt at a Solution

 
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You should show some attempt at the solution. Without that we are not permitted to help.
What about actually citing relevant equations and substituing them?
 
magwas said:
You should show some attempt at the solution. Without that we are not permitted to help.
What about actually citing relevant equations and substituing them?

What I tried was:

ax = 3t/4
vx = -3 + 3t2/4
Fx = ma = 3t/4
Px = Fx vx = -9t/4 + 9t3/32

ay = -1 - 3t/4
vy = -t + 3t2/8 +4
Fy = ma = -1 - 3t/4
Py = Fy vy = 9t3/32 + 9t2/8 - 2t - 4

az = -10
vz = -10t
Fz = ma = -10
Pz = Fz vz = 100t

P = Px + Py + Pz = -9t/4 + 9t3/32 + 9t3/32 + 9t2/8 - 2t - 4 + 100t = 9t3/16 + 9t2/8 + 383t/4 - 4

W = \int P dt

W = 9t4/64 + 3t3/8 + 383t2/8 - 4t

Work Done form 0 sec to 4 sec = W(4)-W(0) = 9(4)4/64 + 3(4)3/8 + 383(4)2/8 - 4(4) = 36 + 24 + 766 - 16 = 810 = 90(9)

Hence K=9[\b]


Please tell if this is correct...
 
you made a typo in vx, however your Px is okay.

Your calculation is otherwise correct.
 
Oh yes, it should be vx = -3 + 3t2/8

But is there any smarter method which is less vulnerable to calculation errors?
 
I used to avoid calculation errors by using a math package and always denoting the units.
Your example in sympy:
Code:
$ isympy 
Python 2.6.4 console for SymPy 0.7.0-git

These commands were executed:
>>> from __future__ import division
>>> from sympy import *
>>> x, y, z = symbols('xyz')
>>> k, m, n = symbols('kmn', integer=True)
>>> f, g, h = map(Function, 'fgh')

Documentation can be found at http://sympy.org/

In [1]: s,m,kg=symbols("s,m,kg",real=True,positive=True)

In [2]: v0=Matrix([-3*m/s,4*m/s,0*m/s])

In [3]: t=symbols("t",real=True,positive=True)

In [4]: a=Matrix([3*t/4*m/s**3,-1.0*m/s**2-3*t/4*m/s**3,-10.0*m/s**2])

In [5]: v=v0+integrate(a,(t,0,t))

In [6]: F=a*1*kg
In [7]: a
Out[7]: 
⎡   3⋅m⋅t    ⎤
⎢   ─────    ⎥
⎢       3    ⎥
⎢    4⋅s     ⎥
⎢            ⎥
⎢  3⋅m⋅t   m ⎥
⎢- ───── - ──⎥
⎢      3    2⎥
⎢   4⋅s    s ⎥
⎢            ⎥
⎢  -10.0⋅m   ⎥
⎢  ───────   ⎥
⎢      2     ⎥
⎣     s      ⎦

In [8]: v
Out[8]: 
⎡               2  ⎤
⎢    3⋅m   3⋅m⋅t   ⎥
⎢  - ─── + ──────  ⎥
⎢     s        3   ⎥
⎢           8⋅s    ⎥
⎢                  ⎥
⎢                 2⎥
⎢4⋅m   m⋅t   3⋅m⋅t ⎥
⎢─── - ─── - ──────⎥
⎢ s      2       3 ⎥
⎢       s     8⋅s  ⎥
⎢                  ⎥
⎢    -10.0⋅m⋅t     ⎥
⎢    ─────────     ⎥
⎢         2        ⎥
⎣        s         ⎦

In [9]: F
Out[9]: 
⎡    3⋅kg⋅m⋅t     ⎤
⎢    ────────     ⎥
⎢         3       ⎥
⎢      4⋅s        ⎥
⎢                 ⎥
⎢   ⎛  3⋅m⋅t   m ⎞⎥
⎢kg⋅⎜- ───── - ──⎟⎥
⎢   ⎜      3    2⎟⎥
⎢   ⎝   4⋅s    s ⎠⎥
⎢                 ⎥
⎢   -10.0⋅kg⋅m    ⎥
⎢   ──────────    ⎥
⎢        2        ⎥
⎣       s         ⎦

In [10]: pp=[]

In [11]: for i in range(3):  pp.append((F[i]*v[i]).expand())
   ...: 

In [12]: P=Matrix(pp)
In [13]: P
Out[13]: 
⎡                        2         2  3               ⎤
⎢                9⋅kg⋅t⋅m    9⋅kg⋅m ⋅t                ⎥
⎢              - ───────── + ──────────               ⎥
⎢                      4           6                  ⎥
⎢                   4⋅s        32⋅s                   ⎥
⎢                                                     ⎥
⎢          2           2             2  2         2  3⎥
⎢  2⋅kg⋅t⋅m    4.0⋅kg⋅m    1.125⋅kg⋅m ⋅t    9⋅kg⋅m ⋅t ⎥
⎢- ───────── - ───────── + ────────────── + ──────────⎥
⎢       4           3             5               6   ⎥
⎢      s           s             s            32⋅s    ⎥
⎢                                                     ⎥
⎢                                2                    ⎥
⎢                    100.0⋅kg⋅t⋅m                     ⎥
⎢                    ─────────────                    ⎥
⎢                           4                         ⎥
⎣                          s                          ⎦

In [14]: P_sum=P[0]+P[1]+P[2]

In [15]: P_sum
Out[15]: 
            2           2             2  2         2  3
95.75⋅kg⋅t⋅m    4.0⋅kg⋅m    1.125⋅kg⋅m ⋅t    9⋅kg⋅m ⋅t 
───────────── - ───────── + ────────────── + ──────────
       4             3             5               6   
      s             s             s            16⋅s    

In [16]: W=integrate(P_sum,t)

In [17]: W
Out[17]: 
            2              2  2             2  3         2  4
  4.0⋅kg⋅t⋅m    47.875⋅kg⋅m ⋅t    0.375⋅kg⋅m ⋅t    9⋅kg⋅m ⋅t 
- ─────────── + ─────────────── + ────────────── + ──────────
        3               4                5               6   
       s               s                s            64⋅s    

In [18]: Wsum=integrate(P_sum,(t,0*s,4*s))

In [19]: Wsum
Out[19]: 
          2
810.0⋅kg⋅m 
───────────
      2    
     s
 
but i can't really use computer during exams!
 
particlemania said:
but i can't really use computer during exams!

You can check units also by hand.
And you can practice.
 

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