# How Much Work Does Ryan Do Lifting and Carrying Potatoes?

• LonelyElectron
In summary, Question 1: Ryan needs to lift a 20 kg bag of potatoes to his shoulder, 1.7 m above the floor, and how much work he does as he carries this bag of potatoes to his car, which is parked 45 m away. The Attempt at a Solution: F_g = mgW = Fdv_f = v_i +atd=1/2(v_i + v_f)tP=W/tQuestion 2: For a 1480 kg car that accelerates uniformly from a position of rest to a speed of 95 km/h in 6.0 s:-calculate the car
LonelyElectron
Hey all! I have three questions below and was hoping someone could proof read, and offer some guidance for the last one... Any help is greatly appreciated!

Question 1:
1. Homework Statement

Ryan needs to carry a bag of potatoes to the car. Determine how much work Ryan has to do to lift a 20 kg bag of potatoes to his shoulder, 1.7 m above the floor, and how much work he does as he carries this bag of potatoes to his car, which is parked 45 m away? Show your work.

F_g = mg
W = Fd

Question 2:

## Homework Statement

For a 1480 kg car that accelerates uniformly from a position of rest to a speed of 95 km/h in 6.0 s:
-calculate the car's acceleration
-determine how far the car traveled during this acceleration
-determine how much work was done by the car during this 6.0 s
-identify what power was developed during this process

## Homework Equations

F=ma
W=fd
v_f = v_i +at
d=1/2(v_i + v_f)t
P=W/t

Question 3:

## Homework Statement

Fill in the empty columns of the following table.​

## Homework Equations

No idea. I used common sense.​

## The Attempt at a Solution

very lost as to how I can fill out the last column...

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Power is expressed in Watts or Joules/s. It is the rate at which energy is generated or dissipated. In short, divide energy by time and you have power. Mind your units. Also, for future reference, please post unrelated questions on separate threads.

LonelyElectron
LonelyElectron said:
very lost as to how I can fill out the last column...
Well, you have kilowatt-hours per week. What is the relationship between kilowatt-hours and work?

LonelyElectron
Question 1 looks good.
Question 2, you did the first two right, but then when calculating the work you multiplied by the distance instead of the acceleration. So your calculation of the work done and the power are both wrong.

In the last column of the table, you have how much energy is used in one week. Since power is energy/time, you just need to divide the energy used in one week by the number of hours in a week.

LonelyElectron
Oh wow... Thanks for the fast replies! Okay, so for question 3, the power is the energy over time. I did the first question in the table and got P=8.62/168=0.051. That's kW because I had kWh as my power, right? So to make it into watts for the answer, I multiply by 1000, which gives me 51 W. Is that correct? As long as the process is right I think I can confidently get the rest :)

As for question 2, I am a little confused by the need for acceleration? The only equation I have been given for work thus far is W=Fd, or W =mgd. Am I missing something?

LonelyElectron said:
As for question 2, I am a little confused by the need for acceleration?
In question 2 you say F = ma but you multiply the mass by the 79.17 which is the number for the distance traveled, not the acceleration.

On edit: kW h/week is not an energy unit; kW h is.

LonelyElectron
kuruman said:
In question 2 you say F = ma but you multiply the mass by the 79.17 which is the number for the distance traveled, not the acceleration.

On edit: kW h/week is not an energy unit; kW h is.

Brilliant on your part! I can't believe I switched my values like that in number 2... Here is the updated. Look better?

As for kWh, not sure I understand. Physics doesn't seem to be my forte...

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Problem 2 agrees with my numbers. To complete the picture, note that you didn't have to go through the acceleration to get the answer. I the car acquires kinetic energy K = ½mv2 in time t, then the average power is P = (½mv2)/t. You get the same number.
LonelyElectron said:
As for kWh, not sure I understand. Physics doesn't seem to be my forte...
You don't really need to understand physics in order to understand kWh. If you understand money, it's enough. Say your weekly salary is $1,000.00. That's a rate of 1 k$/week. If someone asked you how much you make in a month, you would multiply the rate by the number of weeks in a month (= 4) and say (1 k$/week)*4 (week) = 4 k$. If, instead, you were asked how much you make in one working hour (1/40 of a working week), you would say (1 k$/week)*(1/40) week= 0.025 k$. Same thing with kW. The kWh is the amount of energy generated/consumed in one hour. It's an energy and therefore can be expressed in Joules by noting that there are 3600 seconds in one hour. Thus, 1 KWh = 1000 Wh =1000 (J/s)h = 1000(J/s)3600s = 3.6×106 J = 3.6 MJ. So the energy cost of \$0.06 per kWh means that one is paying only 6 cents to buy 3.6 million Joules or 600,000 J for a penny. What a deal!

## 1. What is energy?

Energy is the ability to do work or cause change. It can take many forms, such as light, heat, electricity, and motion.

## 2. How is energy measured?

Energy is measured in joules (J) or any of its derived units, such as kilojoules (kJ) or megajoules (MJ). It can also be measured in calories (cal) or British thermal units (BTU).

## 3. What is the difference between potential and kinetic energy?

Potential energy is energy that is stored and has the potential to do work, while kinetic energy is the energy of motion. An object has potential energy when it is at rest, and it has kinetic energy when it is in motion.

## 4. How is energy transferred or transformed?

Energy can be transferred from one object to another through various means such as heat, electricity, or mechanical work. It can also be transformed from one form to another, such as when chemical energy is converted to electrical energy in a battery.

## 5. Why is energy conservation important?

Energy conservation is important because it helps reduce our carbon footprint and reliance on non-renewable energy sources. It also helps to save money on energy bills and promotes sustainability for future generations.

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