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solving for variables |
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| Aug12-05, 02:42 PM | #1 |
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solving for variables
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? |
| Aug12-05, 02:47 PM | #2 |
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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 |
| Aug12-05, 05:09 PM | #3 |
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Recognitions:
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[tex] v_f^2 - v_i^2 = 2as[/tex] ? |
| Aug12-05, 05:17 PM | #4 |
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solving for variables
the s is still in the equation though
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| Aug12-05, 05:18 PM | #5 |
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Recognitions:
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The m doesn't get squared. |
| Aug12-05, 05:19 PM | #6 |
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Recognitions:
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| Aug12-05, 05:21 PM | #7 |
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oo thats right stupid parentheses.. but i still cant get the first part?
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| Aug12-05, 05:24 PM | #8 |
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Recognitions:
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W = Fs = Mas
vf² - vi² = 2as ==> as = ½(vf² - vi²) So, W = Mas = M*½(vf² - vi²) W = (M/2)(vf² - vi²) ================ |
| Aug12-05, 05:32 PM | #9 |
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o wow yea that makes alot more sense than what i had, i used the wrong eq. lol thank youuu!!
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