Find Force Acting on 5.0kg Stone in 20m Distance

  • Thread starter Thread starter StephenDoty
  • Start date Start date
  • Tags Tags
    Force
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
Sign up Log in
3 replies · 3K views
StephenDoty
Messages
261
Reaction score
0
A 5.0-kg stone is pushed out onto a level horizontal surface with a speed of 2.0 m/s. If
it comes to rest after moving a distance of 20 m, the average frictional force acting on
the block must have been...?

Since it slows down the fk is > than the Fpush.

Fnet= fk-fpush
ma = umg - fpush
ma + fpush = fk

vf^2= v0^2 + 2as
0= 4 + 40a
-4/40 = a = -.1m/s/s

So the stone is slowing down .1m/s every second which will take 20s.

Since the a = -.1m/s/s and f=ma
then wouldn't the fk = 5*-.1= -.5N
Or would you do fk=mg= 5*9.8= 49N

Any help would be great. Thank you.

Stephen Doty
 
Physics news on Phys.org

Discussion

StephenDoty said:
Since it slows down the fk is > than the Fpush.

Fnet= fk-fpush
ma = umg - fpush
ma + fpush = fk
No. The push that set the stone moving is irrelevant. All that matters is that somehow the stone is moving at the given speed and begins to slow down. The only force acting on the stone is friction.

vf^2= v0^2 + 2as
0= 4 + 40a
-4/40 = a = -.1m/s/s
This is good!

What must the force equal to produce this acceleration?
 
Right. The average friction force is 0.5 N.

(If you are familiar with energy methods, you can solve this problem that way as well.)