To calculate the speed of the 6.00 kg block, we can use the principle of conservation of energy. Initially, the block has potential energy due to its position and no kinetic energy. As it falls, its potential energy is converted to kinetic energy. At the bottom of its fall, all of the potential energy has been converted to kinetic energy.
The potential energy of the block at the top of its fall is given by:
PEi = mgh
where m is the mass of the block, g is the acceleration due to gravity, and h is the height of the fall. Plugging in the given values, we get:
PEi = (6.00 kg)(9.81 m/s^2)(1.50 m) = 88.29 J
At the bottom of the fall, all of the potential energy has been converted to kinetic energy:
KEf = 1/2mv^2
where v is the velocity of the block at the bottom of the fall. Equating the two expressions for energy, we get:
PEi = KEf
(6.00 kg)(9.81 m/s^2)(1.50 m) = 1/2(6.00 kg)v^2
Solving for v, we get:
v = sqrt[(2(6.00 kg)(9.81 m/s^2)(1.50 m))/6.00 kg] = 7.67 m/s
Therefore, the speed of the 6.00 kg block after it has descended 1.50 m is 7.67 m/s.