Final answer:
The least acceleration the monkey must have to lift the package off the ground is found by setting the tension in the rope equal for the monkey and the package and solving for the acceleration, which yields 26.4 m/s².
Step-by-step explanation:
To find the least acceleration the monkey must have to lift the package off the ground, we need to apply Newton's second law to the system of the monkey, the package, and the rope and pulley. The forces acting on the monkey and the package include their weights, and the tension in the rope, which is the same on both sides due to it being massless and the pulley being frictionless.
Let a be the least acceleration of the monkey upwards. Then, the package will experience an upward acceleration of a as well. According to Newton's second law, the net force on the monkey is Fnet,monkey = mmonkey × a, and for the package, it's Fnet,package = mpackage × a. As we're looking for the least acceleration to just lift the package, at that instant, the package is in equilibrium with zero net force, thus T - mpackageg = 0.
So, using the relation T = mpackageg for the package and T = mmonkeyg + mmonkeya for the monkey, we set them equal to solve for a:
T = mmonkeyg + mmonkeya = mpackageg, hence
mmonkeya = (mpackage - mmonkey)g, so
a = ((mpackage - mmonkey)g) / mmonkey. Plugging in the values:
a = ((24.0 kg - 6.6 kg) × 10 m/s²) / 6.6 kg = 26.4 m/s².