Newtons 2nd Law & Kinetics pt. 2

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Homework Help Overview

The discussion revolves around a physics problem involving Newton's second law, specifically focusing on a 500kg crate being pushed with a 2000N force while experiencing a 1900N frictional force. Participants are tasked with determining the crate's speed and distance traveled after 10 seconds, as well as calculating the net force acting on it.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the calculation of net force by considering the opposing frictional force and question the initial approach of multiplying mass by time to derive a force. There is discussion about the direction of forces and how to properly sum them as vectors.

Discussion Status

The conversation has progressed with participants clarifying their understanding of force vectors and net force calculations. Some have offered guidance on how to approach the problem using the equation for acceleration, while others express uncertainty about their calculations and the concepts involved.

Contextual Notes

Participants note the absence of the coefficient of friction, which affects the ability to calculate the frictional force directly. There is also a mention of rounding gravitational acceleration for calculations.

jtwitty
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Homework Statement



A 500kg crate is pushed with a 2000N force. The force of the sliding friction acting on the crate is 1900N. How fast will the crate be moving after 10 seconds? How far will it have moved in this time? Determine the net force


Homework Equations



a = [tex]\Sigma[/tex]F/m



The Attempt at a Solution



I tried to find the new force by multiplying 500kg by 10 to get 5000N. Subtracted 5000n frmo 2000n. i don't think negative numbers can be a total force?

2000N-1900N = 100N. ? I don't know what that's used for
 
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jtwitty said:

Homework Statement



A 500kg crate is pushed with a 2000N force. The force of the sliding friction acting on the crate is 1900N. How fast will the crate be moving after 10 seconds? How far will it have moved in this time? Determine the net force


Homework Equations



a = [tex]\Sigma[/tex]F/m



The Attempt at a Solution



I tried to find the new force by multiplying 500kg by 10 to get 5000N. Subtracted 5000n frmo 2000n. i don't think negative numbers can be a total force?

2000N-1900N = 100N. ? I don't know what that's used for

Why would you multiple mass by time to get a force?

You are given the two forces on the crate -- what is the "sum" of those forces? Be sure to include direction in the "sum".

And use your equation that you wrote a = [tex]\Sigma[/tex]F/m

to find the acceleration, which is what you need for the rest of the problem.
 
berkeman said:
Why would you multiple mass by time to get a force?

You are given the two forces on the crate -- what is the "sum" of those forces? Be sure to include direction in the "sum".

And use your equation that you wrote a = [tex]\Sigma[/tex]F/m

to find the acceleration, which is what you need for the rest of the problem.

No i multiplied by 10 (gravity) (we round up frmo 9.8)
 
jtwitty said:
No i multiplied by 10 (gravity) (we round up frmo 9.8)

You are not given the value of the coefficient of friction mu, so you cannot *calculate* the force of friction opposing the pushing motion. Guess you'll just have to figure out that force in a different way...

(Don't let the forest obscure your view of the trees...) :smile:
 
i don't get what you;re saying hahahah

but ok back to the sum of the forces

2000 + 1900? = 3900

idk
 
jtwitty said:
i don't get what you;re saying hahahah

but ok back to the sum of the forces

2000 + 1900? = 3900

idk

Much closer, but remember that the friction force opposes your pushing force. So if you are pushing with a force vector in the +x direction, what direction is the frictional force vector pointing in? And how do you "add" vectors?
 
berkeman said:
Much closer, but remember that the friction force opposes your pushing force. So if you are pushing with a force vector in the +x direction, what direction is the frictional force vector pointing in? And how do you "add" vectors?
idk the word vector :cry:

but i think its pointing in a positive direction?
 
jtwitty said:
idk the word vector :cry:

but i think its pointing in a positive direction?

A vector has magnitude and direction. When two vectors are pointing in opposite directions and you "add" them (like to sum the forces), the resultant vector is the difference in the magnitudes, and points in whichever direction the bigger vector pointed in the first place.

So if you are pushing with 2000N in the +x direction, and the crate's frictional force is pushing in the -x direction with 1900N, what is the resultant vector's magnitude and direction?

You can see the start of this introductory article if you need more info on how to add forces (vectors):

http://en.wikipedia.org/wiki/Vector_addition#Vector_addition_and_subtraction

.
 
it would be poitning in the positive direction b/c its higher by 100N
 
  • #10
jtwitty said:
it would be poitning in the positive direction b/c its higher by 100N

Bingo! So now you know what the sum of the forces is. Use your equation to calculate the acceleration that results, and then just use the kinematic equations of motion (for constant acceleration) to solve the rest of the problem.
 
  • #11
berkeman said:
Bingo! So now you know what the sum of the forces is. Use your equation to calculate the acceleration that results, and then just use the kinematic equations of motion (for constant acceleration) to solve the rest of the problem.

so r u saying that 100N is the net force?

so a = 100 / 500

a = .2 m/s2
 
  • #12
help?
 
  • #13
jtwitty said:
help?

Help why? You are doing fine.
 
  • #14
berkeman said:
Help why? You are doing fine.

i didn't know if i was right or not :(

was i??
 
  • #15
a = .2m/s
d = 1/2 (.2) (100)
d = 10

a = vf /t
.2 = vf /10
2 = vf

done? :) how'd i do
 
  • #16
jtwitty said:
a = .2m/s
d = 1/2 (.2) (100)
d = 10

a = vf /t
.2 = vf /10
2 = vf

done? :) how'd i do

Looks good to me.
 
  • #17
ty man ty :)
 

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