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A finger pushes a chess piece horizontally to the right with a force of 0.06 N. The piece begins to slide to the right.

The coefficient of friction is 0.5 and the chess piece's mass is 0.02 kg. Find the acceleration of the chess piece.
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3 Answers

 
Hi again, Sierra -- chess piece weighs 0.2N, which is the normal force on a flat board ...
friction is 50% of the normal or 0.1N ... since the pushing force of 0.06N < 0.1N friction,
the chess piece, to me,  seems "parked" ==> no acceleration ... All the best :)

Comments

I think what's not mentioned here is the static friction which would result in a static frictional force < the push force as indicated that the piece "slides to the right"? Hence, non-zero acceleration, and slowing down due to kinetic friction.
You may be correct in this example, Alex ... usually for most surfaces, "stiction" is stronger than sliding friction --  more force is needed to "get it going" than to "keep it going" against friction. My angle is that I don't accept the problem as stated ... a "0.6N or even 0.16N" push would seem more "real-world" for me :)
Sum of the forces in the x direction = Force of finger + Force due to friction = ma
 
Force of finger is +0.06N. (Choosing positive direction to the right)
 
Force due to friction is -0.5 * Normal Force (where friction force is resisting and opposite the motion and Normal Force is equal to mg) = -0.098N (The Normal Force = mg because sum of the forces in the y-direction = 0.  Hence: Normal Force + mg = 0 => Normal Force = -m(-g) = mg )
 
Sum of Forces (x-direction) = +0.06N - 0.098N = ma = 0.02kg * a => solve for a
 
a = -1.9m/s^2
 
 This means that as the finger pushes the piece to the right, friction is resisting movement so much that the piece is SLOWING down at a rate of 1.9m/s^2.  Eventually the piece will move to a square and stop.  Checkmate!
 
Force of friction =applied force * coefficient of friction = 0.06 N * 0.5 = 0.03 N
 
acceleration = force/mass = (0.03 N)/(0.02 kg) = 1.5 m/s^2 

Comments

Force due to friction is u*Normal Force.  The question IS ambiguous though.  Is it sliding to the right on its own? Or is it sliding to the right because of the finger?  I think it's the latter.