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
You might be interested in
It is possible to demonstrate that the maximum distance occurs when the angle at which the projectile is fired is 
.
In fact, the laws of motions on both x- and y- directions are
From the second equation, we get the time t at which the projectile hits the ground, by requiring
, and we get:
And inserting this value into Sx(t), we find
And this value is maximum when
, so this is the angle at which the projectile reaches its maximum distance.
So now we can take again the law of motion on the x-axis
And by using
, we find the value of the initial velocity v0:
In fact, the laws of motions on both x- and y- directions are
From the second equation, we get the time t at which the projectile hits the ground, by requiring
And inserting this value into Sx(t), we find
And this value is maximum when
So now we can take again the law of motion on the x-axis
And by using