solving for variables


by marissa12
Tags: solving, variables
marissa12
marissa12 is offline
#1
Aug12-05, 02:42 PM
P: 11
a particle of mass m moving in the x direction at constant acceleration a . During a certain interval of time, the particle accelerates from v_initial to vfinal , undergoing displacement s given by s=x_final-x_initial.

Express the acceleration in terms of v_initial, v_final, and s:
that was easy.. i got

a=(v_final-v_initial)/(s/((v_initial+v_final)/2))

since F=ma
W=F*s

the question is Give an expression for the work in terms of m, v_initial, v_final ?

and i cant seem to eliminate the 's' because i end up getting

W=m*a(see above) *s and i cant get eliminate it? any ideas?
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marissa12
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#2
Aug12-05, 02:47 PM
P: 11
the 2nd part of the problem is w= intergral of { mv*dv} between v_final and v_initial
and we are supposed to express that in m_vinitial and v_final

and i got the answer ((m*v_final)^2-(m*v_initial)^2)/2 and that didnt work. i used the basic intergral formula

fint{t*dt between a and b} = b^2-a^2/t
Fermat
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#3
Aug12-05, 05:09 PM
HW Helper
P: 879
Quote Quote by marissa12
...
the question is Give an expression for the work in terms of m, v_initial, v_final ?

and i cant seem to eliminate the 's' because i end up getting

W=m*a(see above) *s and i cant get eliminate it? any ideas?
can't you use ,

[tex] v_f^2 - v_i^2 = 2as[/tex] ?

marissa12
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#4
Aug12-05, 05:17 PM
P: 11

solving for variables


the s is still in the equation though
Fermat
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#5
Aug12-05, 05:18 PM
HW Helper
P: 879
Quote Quote by marissa12
the 2nd part of the problem is w= intergral of { mv*dv} between v_final and v_initial
and we are supposed to express that in m_vinitial and v_final

and i got the answer ((m*v_final)^2-(m*v_initial)^2)/2 and that didnt work. i used the basic intergral formula

fint{t*dt between a and b} = b^2-a^2/t
[tex]\int_{vi}^{vf} mv dv = m[v^2/2]_{vi}^{vf}[/tex]

The m doesn't get squared.
Fermat
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#6
Aug12-05, 05:19 PM
HW Helper
P: 879
Quote Quote by marissa12
the s is still in the equation though
substitute for "as" from one eqn into t'other.
marissa12
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#7
Aug12-05, 05:21 PM
P: 11
oo thats right stupid parentheses.. but i still cant get the first part?
Fermat
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#8
Aug12-05, 05:24 PM
HW Helper
P: 879
W = Fs = Mas

vf² - vi² = 2as ==> as = ½(vf² - vi²)

So,

W = Mas = M*½(vf² - vi²)
W = (M/2)(vf² - vi²)
================
ixbethxi
ixbethxi is offline
#9
Aug12-05, 05:32 PM
P: 13
o wow yea that makes alot more sense than what i had, i used the wrong eq. lol thank youuu!!


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