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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.
The work done by the battery is equal to the charge transferred during the process times the potential difference between the two terminals of the battery:
where q is the charge and
is the potential difference.
In our problem, the work done is W=39 J while the potential difference of the battery is
, so we can find the charge transferred by the battery:
where q is the charge and
In our problem, the work done is W=39 J while the potential difference of the battery is