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

## 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.

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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 :)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.

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