Question

An elevator and counterweight are like Atwood’s machine. An elevator, M=100kg, has a counter weight m=90kg connected by a cable over a massless pulley with no friction. The elevator falls, starting from rest, a distance 20.5 m and lands. What the final kinetic energy of the system, in J, just before the elevator lands?

Answer 1

v = 6.34 m / s

Step-by-step explanation:

For this exercise let's use the concept of mechanical energy, write the energy of the system (elevator plus counterweight) at two points

Initial. Elevator highest point

The elevator mass is M and that of the counterweight m

Em₀ = U = M g h + mg (0)

The counterweight is on the floor

Final. elevated at the lowest point

`Em_(f)` = K + U = ½ M v² + Mg (0) + m g h

as there is no friction

`Em_(f)` = Em₀

½ M v² + m g h = M g h

v² = (M-m) gh 2 / M

v = √ [(100-90) 9.8 20.5 2/100]

v = √ 40.18

v = 6.34 m / s