Calculating K in the XYZ Space: A Homework Challenge

AI Thread Summary
The discussion revolves around calculating the value of K in a physics homework problem involving a particle projected in XYZ space under gravity. The initial conditions include a mass of 1 kg, an initial velocity, and specific acceleration components. The total work done over a time interval of 0 to 4 seconds is given as 90K Joules, leading to the conclusion that K equals 9 after performing the necessary calculations. Participants suggest using computational tools to minimize calculation errors, while also emphasizing the importance of practicing manual calculations for exams. The final consensus confirms that K=9 is correct based on the calculations provided.
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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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