A passenger elevator consists of an elevator cage of 900 kg (empty) and a counterweight of 990 kg connected by a cable running over a pair of pulleys (Figure 6.9). Neglect the masses of the cable and of the pulleys. If the elevator carries four passengers of 70 kg each, what speed will the elevator attain running down freely from a height of 10 m, starting from rest?

I was trying using this formula of aceleration $$a = (m2-m1/m1+m2)g$$

then using eq $$x0+v_0t+1/2at^2$$

solvign for t gives me $$t = \sqrt{2/a}$$

once we have t we can use eq $$v= v_o-gt$$

however this didnt work out

  • $\begingroup$ It will simpler to just use work-energy. V1+T1 = V2 + T2 (there are no external forces). T1=0 since starting from rest. So all what you have is to find V1,V2 using (mgh) and hence you can find T2 which gives you the final speed. $\endgroup$
    – Nasser
    Jan 23 at 0:05
  • $\begingroup$ Thanks for the help Nasser however in my class we haven´t reach work and energy $\endgroup$ Jan 23 at 0:34