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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Answer:
The arrow is at a height of 500 feet at time t = 2.35 seconds.
Explanation:
It is given that,
An arrow is shot vertically upward at a rate of 250 ft/s, v₀ = 250 ft/s
The projectile formula is given by :
We need to find the time(s), in seconds, the arrow is at a height of 500 ft. So,
On solving the above quadratic equation, we get the value of t as, t = 2.35 seconds
So, the arrow is at a height of 500 feet at time t = 2.35 seconds. Hence, this is the required solution.