What is the required force to accelerate a child on a sled at 1.12 m/s2?

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Homework Help Overview

The discussion revolves around calculating the force required to accelerate a child on a sled with a specified mass and acceleration. Additionally, there are related questions about horizontal forces needed to move objects with and without friction.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of Newton's second law (F=MA) and question the calculations related to the required force for the sled. There are also inquiries about the effects of friction on the force needed to maintain constant speed.

Discussion Status

There is ongoing dialogue about the calculations and interpretations of the problems presented. Some participants express uncertainty about their answers, while others provide insights into the relationship between force, mass, and acceleration. The conversation reflects a mix of attempts to clarify concepts and resolve discrepancies in calculations.

Contextual Notes

Participants mention specific values for mass and acceleration, as well as coefficients of friction, but there is a noted lack of consensus on the correct calculations and interpretations of the problems. Some participants express urgency for answers due to time constraints.

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Physics question, need help A.S.A.P.!

What force is needed to accelerate a child on a sled (total mass = 61.0 kg) at 1.12 m/s2?
 
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I got another one,

If the coefficient of kinetic friction between a 37 kg crate and the floor is 0.25, what horizontal force is required to move the crate at a steady speed across the floor?

What horizontal force is required if µk is zero?
 
FORCE = MASS x Acceleration.
 
Rapta3 said:
I got another one,

If the coefficient of kinetic friction between a 37 kg crate and the floor is 0.25, what horizontal force is required to move the crate at a steady speed across the floor?

What horizontal force is required if µk is zero?

to move it at a steady speed what is the sum of the forces? Another way of putting this question is what is the acceleration on the sled is the speed is CONSTANT

Once you figure that out what is the sum of the forces (applied and friction)?
 
I need answers guys, i appreciate you trying to help me figure it out but I have no time, I have other things to do so please help me if you can, I would greatly appreciate it, thanks.
 
Is 73.2 N the correct answer for my first question?
 
Help! Cmon!
 
Yes you're right on the first one. For the second one, the force required is = [tex]\mu_s m g[/tex]
 
Ok help me out then here, i know F=MA but for the first question i posted, i did that and the answer got was wrong so I don't know how to get the right answer...
 
  • #10
Friction force neglected, the force F can accelerate a child/sled of mass M to an acceleration A. I don't see how you could be wrong unless there's something in the question you didn't mention
 
  • #11
No that's the complete question...
 
  • #12
I didn't do the calculation but it looked right. You're off by 5N. 1.12*61 != 73.2
 
  • #13
thats what i said, look above i got 73.2
 
Last edited:
  • #14
Rapta3 said:
thats why i said, look above i got 73.2

I help with math and physics, not arithmetic, sorry :(
 
  • #15
Ok someone help me with my second question, please
 
  • #16
To move an object at a steady speed across a surface with friction you must apply a force equal to the friction. The same applies for the second part except there is no friction therefore you don't need any force.
 
  • #17
Ok thank you.
 
  • #18
If the coefficient of kinetic friction between a 39 kg crate and the floor is 0.25, what horizontal force is required to move the crate at a steady speed across the floor?


If it is what you say it is, it should be .25, but its not so what else is missing?
 
  • #19
The force of friction is equal to [tex]\mu_k mg[/tex] not just the coefficient of friction. Hope this helps.
 
  • #20
Ok well thanks for everything... time for some sleep, ill talk to guys later, bye.
 
  • #21
You're welcome. Glad to hear we helped you out some.
 

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