# 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.

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

Last edited by a moderator: May 3, 2017
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