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

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To accelerate a child on a sled with a total mass of 61.0 kg at 1.12 m/s², the required force is calculated using the formula F = MA, resulting in 68.32 N, not 73.2 N as initially thought. For moving a crate with a mass of 37 kg at a steady speed across a surface with a coefficient of kinetic friction of 0.25, the force needed equals the friction force, which is calculated as F = µ_k * m * g. If there is no friction (µ_k = 0), no force is required to maintain steady speed. The discussion also highlights the importance of correctly applying the formulas for force and friction. Overall, the participants provided clarification on the calculations and concepts involved in these physics problems.
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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 = \mu_s m g
 
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 \mu_k mg 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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