Physics hwk; applied force

1. Nov 2, 2007

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

2. Nov 2, 2007

hage567

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

3. Nov 2, 2007

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

4. Nov 2, 2007

hage567

Do you know the equation for the frictional force?

5. Nov 2, 2007

vince_lu

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

6. Nov 2, 2007

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

7. Nov 2, 2007

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]

8. Nov 2, 2007

hage567

Yes, that's right!

9. Nov 2, 2007

vince_lu

YYYesssss THANKYOU

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