To solve this problem we will apply the concepts of equilibrium and Newton's second law.
According to the description given, it is under constant ascending acceleration, and the balance of the forces corresponding to the tension of the rope and the weight of the elevator must be equal to said acceleration. So


Here,
T = Tension
m = Mass
g = Gravitational Acceleration
a = Acceleration (upward)
Rearranging to find T,



Therefore the tension force in the cable is 10290.15N
Answer: angular acceleration = 
Given:
Distance from center of axis = 1.6 m
Time taken to complete one revolution = 4.7 sec
Therefore, we can evaluate the angular acceleration using the following formula:



= 
Answer:
3.62m/s and 2.83m/s
Explanation:
Apply conservation of momentum
For vertical component,
Pfy = Piy
m* Vof (sin38) - m*Vgf (sin52) = 0
Divide through by m
Vof(sin38) - Vgf(sin52) = 0
Vof(sin38) = Vgf(sin52)
Vof (sin38/sin52) = Vgf
0.7813Vof = Vgf
For horizontal component
Pxf= Pxi
m* Vof (cos38) - m*Vgf (cos52) = m*4.6
Divide through by m
Vof(cos38) + Vgf(cos52) = 4.6
Recall that
0.7813Vof = Vgf
Vof(cos38) + 0.7813 Vof(cos52) = 4.6
0.7880Vof + 0.4810Vof = 4.
1.269Vof = 4.6
Vof = 4.6/1.269
Vof = 3.62m/s
Recall that
0.7813Vof = Vgf
Vgf = 0.7813 * 3.62
Vgf = 2.83m/s
To determine the displacement, since we are given the potential energy, we use the equation for potential energy. For a spring, it is one-half the product of the spring constant and the square of the displacement. We do as follows:
PE = kx^2/2
5 Nm = 50N/m (x^2)
x = 0.32 m
Therefore, the displacement would be 0.32 m.