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?

v = 6.34 m / s

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

    = K + U = ½ M v² + Mg (0) + m g h

as there is no friction

    = 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

a small object of mass 3.66 g and charge -17.6 µc is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. what are the magnitude and direction of the electric field?

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