A person pulls a bucket of water up from a well with a rope. Assume the initial and final speeds of the bucket are zero (Vi-Vf-0
), and that the person lifted the bucket a vertical distance h. By looking at energy changes in the bucket-Earth physical system, we can make sense of the force the person must exert to pull the bucket up and the amount of work the person does. a) Think about whether this is an open or closed system. What energy systems undergo a change in energy during the process? Construct a particular Energy-System Diagram for this process and include the algebraic expression of energy conservation (in terms of AE's and if open, any Heat or Work). b) Substitute the algebraic expression we use for the change in gravitational potential energy and solve algebraically for the work done by the person on the rope/bucket. c) Use the definition of work in terms of force and distance (see p.39-40 of Chapter 2 in the Course Notes) to find the average force exerted by the person on the rope and bucket while they are lifting it up.
1 answer:
Answer:
a
This a closed system because the mass of the system is conserved
The energy system that undergoes change is the Potential energy system
The energy system diagram is shown on the first uploaded image
b
Work done = Change in gravitational potential energy
So solving algebraically for work done would be
Work done =
where m is mass
g is acceleration due to gravity
and h is the height
c
Work done in terms of force and distance is = mg
where m is mass of bucket and
g is acceleration due to gravity
Explanation:
a) At the start, potential and kinetic energy were zero. so, energy is zero.
As the person pulls the bucket up, the potential energy becomes mgh.
so,final energy will be consisting of only potential energy.
B) Here work done is equal to change in gravitational potential energy.
W =
W = m*g*h
where g = 9.9 m/
C) Work = force * distance
mgh = force * h
force = mg
force = weight of bucket
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The question is incomplete. The complete question is :
Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a minimum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity increases until reaching a maximum when you are directly in front of one of the speakers. The speed of sound in the room is 340 m/s.
What is the frequency of the sound?
Solution :
Given :
The distance between the two loud speakers,
The speaker are in phase and so the path difference is zero constructive interference occurs.
At the point , the speakers are out of phase and so the path difference is
Therefore,
Thus the frequency is :
Hz
