Part a of the drawing shows a bucket of water suspended from the pulley of a well; the tension in the rope is 92.5 N. Part b shows the same bucket of water being pulled up from the well at a constant velocity. What is the tension in the rope in part b?
b) at constant velocity, zero acceleration, weight mg is being supported.
So i got the equation T=mg...but how do I find m?
To find the tension in the rope in part b, we need to understand the concept of static equilibrium. When the bucket is pulled up at a constant velocity, the forces acting on the bucket are balanced. This means that the net force on the bucket is zero.
In part a, the tension in the rope is 92.5 N when the bucket is in equilibrium. We can assume that the weight of the bucket (the force due to gravity acting on it) is also 92.5 N. This is because, in equilibrium, the tension in the rope should be equal to the weight of the object being balanced.
Now, in part b, when the same bucket is pulled up at a constant velocity, the weight of the bucket remains the same (92.5 N) since the mass of the bucket has not changed. However, in order to maintain the constant velocity, an additional force called the force of pulling is required. This force cancels out exactly the weight of the bucket.
Therefore, the tension in the rope in part b is equal to the force of pulling, which is also 92.5 N.
In summary, the tension in the rope in both parts a and b is 92.5 N.