A10 kg monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 15 kg package on the ground. (a) what is the magnitude of the least acceleration the monk"y must have if it is to lift the package off the ground? if, after the package has been lifted, the monkey stops its climb and holds onto the rope, what are the (b) magnitude and (c) direction of the monkey's acceleration and (d) the tension in the rope?

A. 4,9 m/s2

B. 2,0 m/s2

C. 120 N

Explanation:

In the image, 1 is going to represent the monkey and 2 is going to be the package.  Let a_mín be the minimum acceleration that the monkey should have in the upward direction, so the package is barely lifted. Apply Newton’s second law of motion:

If the package is barely lifted, that means that T=m_2*g; then:

Solving the equation for a_mín, we have:

Once the monkey stops its climb and holds onto the rope, we set the equation of Newton’s second law as it follows:

For the monkey:

For the package:

The acceleration a is the same for both monkey and package, but have opposite directions, this means that when the monkey accelerates upwards, the package does it downwards and vice versa. Therefore, the acceleration a on the equation for the package is negative; however, if we invert the signs on the sum of forces, it has the same effect. To be clearer:

For the package:

We have two unknowns and two equations, so we can proceed. We can match both tensions and have:

Solving a, we have

We can then replace this value of a in one for the sums of force and find the tension T:

MERICA

Explanation:

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