Calculating Work in Terms of Mass and Initial/Final Velocities

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Homework Help Overview

The discussion revolves around calculating work done by a particle of mass m moving with constant acceleration a, transitioning between initial and final velocities. Participants are tasked with expressing work in terms of mass and these velocities, while grappling with the displacement variable.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between work, force, and displacement, attempting to eliminate displacement from their equations. Some suggest using kinematic equations to relate acceleration and displacement, while others question the validity of their integrals and substitutions.

Discussion Status

The discussion is active, with participants sharing various approaches to express work in the desired terms. Some have provided partial expressions and insights, while others are still struggling with the integration and substitution steps. There is a mix of interpretations regarding the correct application of formulas.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can share or the methods they can employ. The presence of conflicting approaches indicates a lack of consensus on the best way to eliminate displacement from the work equation.

marissa12
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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 can't seem to eliminate the 's' because i end up getting

W=m*a(see above) *s and i can't get eliminate it? any ideas?
 
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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
 
marissa12 said:
...
the question is Give an expression for the work in terms of m, v_initial, v_final ?

and i can't seem to eliminate the 's' because i end up getting

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

v_f^2 - v_i^2 = 2as ?
 
the s is still in the equation though
 
marissa12 said:
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
\int_{vi}^{vf} mv dv = m[v^2/2]_{vi}^{vf}

The m doesn't get squared.
 
marissa12 said:
the s is still in the equation though
substitute for "as" from one eqn into t'other.
 
oo that's right stupid parentheses.. but i still can't get the first part?
 
W = Fs = Mas

vf² - vi² = 2as ==> as = ½(vf² - vi²)

So,

W = Mas = M*½(vf² - vi²)
W = (M/2)(vf² - vi²)
================
 
Last edited:
o wow yea that makes a lot more sense than what i had, i used the wrong eq. lol thank youuu!
 

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