For rotational equilibrium of the door we can say that torque due to weight of the door must be counter balanced by the torque of external force
here weight will act at mid point of door so its distance is half of the total distance where force is applied
here we know that
now we will have
so our applied force is 72.5 N
Answer:
The vertical distance is
Explanation:
From the question we are told that
The mass of the cylinder is m
The kinetic frictional force is f
Generally from the work energy theorem
Here E the the energy of the spring which is increasing and this is mathematically represented as
Here k is the spring constant
P is the potential energy of the cylinder which is mathematically represented as
And
is the workdone by friction which is mathematically represented as
So
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