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?
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
W = Fs = Mas vf² - vi² = 2as ==> as = ½(vf² - vi²) So, W = Mas = M*½(vf² - vi²) W = (M/2)(vf² - vi²) ================