Calculating Work in Terms of Mass and Initial/Final Velocities

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SUMMARY

The discussion focuses on deriving the work done by a particle of mass m moving under constant acceleration a, transitioning from an initial velocity v_initial to a final velocity v_final. The key formula derived is W = (m/2)(v_final² - v_initial²), which eliminates displacement s by substituting it with the relationship between velocities and acceleration. Participants clarified the integration process and corrected initial misunderstandings regarding the application of the integral formula for work.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Familiarity with kinematic equations, specifically v_f² - v_i² = 2as
  • Basic knowledge of calculus, particularly integration
  • Ability to manipulate algebraic expressions involving mass and velocity
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  • Study the derivation of work-energy principles in classical mechanics
  • Learn about kinematic equations and their applications in physics
  • Explore integration techniques for calculating work done in variable force scenarios
  • Review the relationship between force, mass, and acceleration in different contexts
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Students of physics, educators teaching mechanics, and anyone interested in understanding the work-energy theorem and its applications in motion analysis.

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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