Answer:
D) 25 m/s
Explanation:
In order to solve this problem we must use the following kinematics equation.
where:
Vf = final speed [m/s]
Vi = initial speed = 0
a = acceleration = 5[m/s^2]
t = time = 5[s]
After 5 seconds the acceleration is equal to 5 [m/s^2]
Now replacing the values in the equation:
Vf = 0 + (5*5)
Vf = 25[m/s]
Answer:
x = 0.176 m
Explanation:
For this exercise we will take the condition of rotational equilibrium, where the reference system is located on the far left and the wire on the far right. We assume that counterclockwise turns are positive.
Let's use trigonometry to decompose the tension
sin 60 = / T
T_{y} = T sin 60
cos 60 = Tₓ / T
Tₓ = T cos 60
we apply the equation
∑ τ = 0
-W L / 2 - w x + T_{y} L = 0
the length of the bar is L = 6m
-Mg 6/2 - m g x + T sin 60 6 = 0
x = (6 T sin 60 - 3 M g) / mg
let's calculate
let's use the maximum tension that resists the cable T = 900 N
x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)
x = (4676 - 5880) / 6860
x = - 0.176 m
Therefore the block can be up to 0.176m to keep the system in balance.