# Work and Energy Test Bank: 5 Questions

• bssmagik
The second law of thermodynamics is the key.In summary, the conversation involves a student asking for help on a set of 5 questions involving work and energy. The student has already confirmed the answers with a key provided by the teacher, but is struggling to understand the equations and how to solve the problems. The expert summarizer suggests that the student should spend more time reading the textbook and provides guidance on how to approach one of the questions.
bssmagik
here's a set of 5 questions: Work and Energy

## Homework Statement

1. While traveling to school at 27 m/s your car runs out of gas 15 km from the nearest gas station. If the station is 16 m above your current elevation, how fast will the car be going when it reaches the gas station? Ignore friction.

2. A lead ball is traveling at 275 m/s when it strikes a steel plate and comes to a stop. If all its kinetic energy is converted into heat and none of the energy leaves the bullet, what is the bullet’s change in temperature?

3. A 637 gram sample of water at 92 C is mixed with 843 grams of water at 27 C. Assume no heat loss to the surroundings. What is the final temperature of the mixture?

4. Tommy throws his car keys straight up into the air from height of 1.5 m above the ground. His keys strike the ground at a speed of 7.7 m/s. Ignoring friction, how fast did Tommy throw his keys?

5. A 1.5 kg bass is hooked by a fisherman. The fisherman plays the bass by allowing the fish to swim off at 2.1 m/s before braking his reel and stopping the bass in 37 cm. How much tension is exerted on the line? Assume the fish is neutrally buoyant.

## Homework Equations

i know that:

(Cwater = 4180 J/kgK)

## The Attempt at a Solution

I did all of these in my head/on a calculator, but I'm not sure how any of the equations work or what they are. I've always been able to get the answer without any knowledge of how i did it. Help??

20.8 m/s
302.5 C
54.98 C
5.47 m/s
8.94 N

feel free to pick just one.
i just really really need some help here!

bssmagik said:
feel free to pick just one.
i just really really need some help here!

If you can't be troubled to write out your solution, we are not going to do the problems just to verify your answers. Why do you think you need help?

I'll pick one. How did you do #2?

well, I've already confirmed the answers with a key the teacher has posted online. but as to getting to the answer, i just performed "random" functions in the calculator until i got to the answers that seemed correct. i have a basic idea...

#2

equations:
Q=mcT
U=Q-W
U=mgh
1/2 mv^2=W

W + U = mcT

1/2 mv^2 + mgh = mcT

(v^2 + h) / 2c =T

assuming the displacement (h) is zero, that works?

Last edited:
bssmagik said:
well, I've already confirmed the answers with a key the teacher has posted online. but as to getting to the answer, i just performed "random" functions in the calculator until i got to the answers that seemed correct. i have a basic idea...

I think you should spend, oh, a few months reading your textbook. You're never going to get anywhere with this "technique."

- Warren

bssmagik said:
well, I've already confirmed the answers with a key the teacher has posted online. but as to getting to the answer, i just performed "random" functions in the calculator until i got to the answers that seemed correct. i have a basic idea...

#2

equations:
Q=mcT
U=Q-W
U=mgh
1/2 mv^2=W

W + U = mcT

1/2 mv^2 + mgh = mcT

(v^2 + h) / 2c =T

assuming the displacement (h) is zero, that works?

Q=mcT <== Is any heat added to the lead in this problem? What is T in this equation? What is c? What value did you use in your computation?

U=Q-W <== What is U and how is it related to T? Is any work done on by the lead (assume the plate does not move). Is any work done on the lead? What is the significance of this for the sign of W in this equation? How do your answers connect work done to changing the temperature?

U=mgh <== Is this U the same as the previous U?

1/2 mv^2=W <== Is this the same W as in the U = Q-W equation?

W + U = mcT <== What if Q = 0? What does this mean?

(v^2 + h) / 2c =T <== If h were not zero this would not have worked. You dropped the g, and lost a factor of 2

Try to answer the questions. They are important to your understanding of how the correct final equation

(v^2)/2c = ΔT

thanks for the help - one last question.

after looking over all the other questions (and reading the book), i figured them out.
but i am completely lost on #3.
like, i have NO idea what to do.

could someone help me reach a working equation?

bssmagik said:
thanks for the help - one last question.

after looking over all the other questions (and reading the book), i figured them out.
but i am completely lost on #3.
like, i have NO idea what to do.

could someone help me reach a working equation?

The correct version of Q=mcT and the fact that heat lost by the water that cools = heat gained by the water that warms up.

## 1. What is work and energy?

Work and energy are two closely related concepts in the field of physics. Work is defined as the transfer of energy from one object to another, while energy is the ability of an object to do work.

## 2. How is work calculated?

Work is calculated by multiplying the force applied to an object by the distance it moves in the direction of the force. This can be represented by the equation W = Fd, where W is work, F is force, and d is distance.

## 3. What are the different forms of energy?

There are several forms of energy, including kinetic energy (energy of motion), potential energy (stored energy), thermal energy (heat), chemical energy (stored in chemical bonds), and nuclear energy (stored in the nucleus of an atom).

## 4. How is energy conserved?

The law of conservation of energy states that energy cannot be created or destroyed, only transferred or transformed from one form to another. This means that the total amount of energy in a closed system remains constant.

## 5. How is energy related to work?

Energy and work are related in that work is a way of transferring energy from one object to another. When work is done on an object, its energy increases, and when work is done by an object, its energy decreases.

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