Answer:
the tangential velocity of the student is 4.89 m/s.
Explanation:
Given;
the radius of the circular path, r = 3.5 m
duration of the motion, t = 4.5 s
let the student's tangential velocity = v
The tangential velocity of the student is calculated as follows;
Therefore, the tangential velocity of the student is 4.89 m/s.
Answer:
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:
