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