Physics Help, Could someone check my work test coming up soon

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Homework Help Overview

The discussion revolves around calculating work done in various physics scenarios, including a parent pushing a stroller, lifting a picture frame, and lowering a textbook. The subject area includes concepts of work, energy, and potential energy in mechanics.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants attempt to calculate work done using force and distance, questioning unit conversions and the correctness of their calculations. Some participants express uncertainty about the application of formulas and the interpretation of potential energy changes.

Discussion Status

There is active engagement with multiple participants checking each other's calculations and questioning assumptions about the problems. Some guidance has been offered regarding unit consistency and the need for clarity in equations, but no consensus has been reached on the correctness of the answers.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can provide or seek. There is also a mention of needing to confirm answers before responding, indicating a desire for accuracy in the discussion.

Ion1776
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1. A parent pushes a baby stroller from home to daycare along a level road with a force of 38 N directed at an angle of 30° below the horizontal. If daycare is 0.83 km from home, how much work is done by the parent?

(38N)(cos(-30))(.83km)=27.21 J

2. You have a 0.750 kg picture frame on a mantle 1.50 m high. How much work was required to lift the frame from the ground to the mantle?

0.750*9.80=7.35N(1.50)=11.025 J

3. You lower a 2.50 kg textbook (remember when textbooks used to be made out of paper instead of being digital?) from a height of 1.81 m to 1.50 m. What is its change in potential energy?

1.81-1.50=.31
(2.50kg)(9.80)(.31)

=7.60 J

Please if someone could confirm these by working them out, please be sure of the answer before responding, Thanks so much
 
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In 1, a Joule is a Newton meter. So check your units...

In 2, you are loose with your one "equation". The LHS does not equal the middle, but it looks like you got the right answer anyway. You should show all terms and units on each side of your equation.

3 looks okay.
 
On #1 Should .83 Km be 830 meters, if that is the case will it make the question correct.

(38N)(cos(-30))(830m)= 27309.4568 J?
 
Need multiple opinions on these physics problems

1. A parent pushes a baby stroller from home to daycare along a level road with a force of 38 N directed at an angle of 30° below the horizontal. If daycare is 0.83 km from home, how much work is done by the parent?

(38N)(cos(-30))(830m)=7309.4568 J?

2. You have a 0.750 kg picture frame on a mantle 1.50 m high. How much work was required to lift the frame from the ground to the mantle?

0.750*9.80=7.35N(1.50)=11.025 J

3. You lower a 2.50 kg textbook (remember when textbooks used to be made out of paper instead of being digital?) from a height of 1.81 m to 1.50 m. What is its change in potential energy?

1.81-1.50=.31
(2.50kg)(9.80)(.31)

=7.60 J

Please if someone could confirm these by working them out, please be sure of the answer before responding, Thanks so much
 


Ion1776 said:
1. A parent pushes a baby stroller from home to daycare along a level road with a force of 38 N directed at an angle of 30° below the horizontal. If daycare is 0.83 km from home, how much work is done by the parent?

(38N)(cos(-30))(830m)=7309.4568 J?
The left hand side is correct, but the right hand side is wrong. You might want to check that you punched the numbers into your calculator correctly
Ion1776 said:
2. You have a 0.750 kg picture frame on a mantle 1.50 m high. How much work was required to lift the frame from the ground to the mantle?

0.750*9.80=7.35N(1.50)=11.025 J
The orange part is correct. You should note that the blue part is not equal to the orange part, so it is incorrect to state that they are.

Ion1776 said:
3. You lower a 2.50 kg textbook (remember when textbooks used to be made out of paper instead of being digital?) from a height of 1.81 m to 1.50 m. What is its change in potential energy?

1.81-1.50=.31
(2.50kg)(9.80)(.31)

=7.60 J
Has the textbook gained, or lost potential energy?
 


are the answers correct for 2 and 3?

On 2 what to with the blue side

lost potential energy on 3 right?
 


The answer to 3 is incorrect. ΔU = mgΔh, where Δh is the change in height, i.e. final value minus initial value.
 


So should the answer be (2.50)(9.80)(-.31)=-7.595
 
(Two threads with the same questions merged into one...)
 
  • #10
So should the answer be (2.50)(9.80)(-.31)=-7.595 FOR #3
 

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