
#1
Aug1205, 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_finalx_initial.
Express the acceleration in terms of v_initial, v_final, and s: that was easy.. i got a=(v_finalv_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? 



#2
Aug1205, 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^2a^2/t 



#3
Aug1205, 05:09 PM

HW Helper
P: 879

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



#4
Aug1205, 05:17 PM

P: 11

solving for variables
the s is still in the equation though




#5
Aug1205, 05:18 PM

HW Helper
P: 879

The m doesn't get squared. 



#6
Aug1205, 05:19 PM

HW Helper
P: 879





#7
Aug1205, 05:21 PM

P: 11

oo thats right stupid parentheses.. but i still cant get the first part?




#8
Aug1205, 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) ================ 



#9
Aug1205, 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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