Calculating Work in Terms of Mass and Initial/Final Velocities

  • Thread starter Thread starter marissa12
  • Start date Start date
  • Tags Tags
    Variables
AI Thread Summary
The discussion focuses on deriving an expression for work in terms of mass and initial and final velocities for a particle undergoing constant acceleration. The initial approach involved using the formula for acceleration and the work-energy principle, but participants struggled to eliminate displacement from the equations. A key insight was to use the kinematic equation relating velocity and displacement, leading to the conclusion that work can be expressed as W = (m/2)(v_final² - v_initial²). This clarified the relationship between mass, velocity, and work, resolving earlier confusion. Ultimately, the correct expression for work was confirmed, emphasizing the importance of proper equation manipulation.
marissa12
Messages
11
Reaction score
0
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?
 
Physics news on Phys.org
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!
 
Back
Top