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

[vince_lu]

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

 

[hage567]

What have you tried? You must show some work in order to get help. What's Newton's second law?

 

[vince_lu]

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

 

[hage567]

Do you know the equation for the frictional force?

 

[vince_lu]

No, I am not sure I think its F=Fnet-Ff

 

[hage567]

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 $$\Sigma F=ma$$

The frictional force is given by $$f = \mu mg$$

See if you can try it now.

Maybe this can help you out too

http://hyperphysics.phy-astr.gsu.edu/hbase/fric.html#fri

 

[vince_lu]

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]

 

[hage567]

Yes, that's right!

 

[vince_lu]

YYYesssss THANKYOU