What is the Applied Force Needed to Accelerate a Curling Stone on Ice?

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Homework Help Overview

The discussion revolves around calculating the applied force needed to accelerate a curling stone on ice, given its mass, acceleration, and the coefficient of kinetic friction. The subject area includes dynamics and frictional forces in physics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of Newton's second law and the setup of equations involving net force and frictional force. There are inquiries about the correct equations to use and how to incorporate the coefficient of friction into the calculations.

Discussion Status

Some participants have provided guidance on the relevant equations and concepts, while others are working through their understanding of the problem. There is a sense of progress as one participant expresses improved comprehension of the topic.

Contextual Notes

Participants are reminded to show their work to receive help, indicating a focus on understanding the problem-solving process rather than just finding the answer.

vince_lu
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a curler exerts a force forward on a 19kg curling stone and gives it an acceleration of 1.8m/s(squared) [forward]. the coefficient of kinetic friction of the ice on the curling stone is 0.080 [back]. calculate the value of the applied force. thank you in advanced
 
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What have you tried? You must show some work in order to get help. What's Newton's second law?
 
Newton's Second law is an object will accelerate only wher there is a net external force acting

So far I figured out:
m=19kg
a(acceleration)=1.8m/s^2[forward]
(u)coefficient of Fk=0.080[back]
F(applied force)=?

Idon't know how to setup the equation
 
Do you know the equation for the frictional force?
 
No, I am not sure I think its F=Fnet-Ff
 
There are two forces acting on the stone. One is the applied force and one is the frictional force. The sum of the two is the net force. Remember this [tex]\Sigma F=ma[/tex]

The frictional force is given by [tex]f = \mu mg[/tex]

See if you can try it now.

Maybe this can help you out too

http://hyperphysics.phy-astr.gsu.edu/hbase/fric.html#fri
 
Last edited by a moderator:
I think I'm getting it better now.
Would it then be: 0.080[back] x 19kg x 9.8 + (19kg)(1.8m/s^2)
F=49N[forward]
 
Yes, that's right!
 
YYYesssss THANKYOU
 

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