Question

A player uses a hockey stick to push a puck at a constant velocity across the ice. The weight of the puck is 170 N. The
coefficient of friction is 0.0600. With what force must the player push the puck so that his force just equals the frictional
force?
28.3 N
0.999 N
0.102 N
0.0104 N

Answer 1

Answer: The minimum force by which player should hit the puck to move it is 0.102 N.

Explanation:

Weight of the puck = 1.70 N

This weight of the puck is the force acting normal to the surface that is:

N = 1.70 N

The coefficient of friction =

Frictional force =

The minimum force by which player should hit the puck to move it is 0.102 N.

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Answer 2

Answer:

F=10.2N : Force with which the player must push the disc

Explanation:

Conceptual analysis

We apply Newton's first law because the speed is constant in horizontal direction (x):

∑Fx = 0 Formula (1)

We apply Newton's first law because the speed is zero in vertical direction (y):

∑Fy= 0 Formula (2)

∑F: algebraic sum of forces in Newton (N)  

We calculate the friction force with the following formula:

Ff=f*N  Formula (3)

Ff= friction force in Newtons (N)

f = coefficient of friction

N= Normal force in Newtons (N)

Known data:

Wp = 170 N  : puck weight

f = 0.0600 :coefficient of friction

Problem development

We replace the data in formula (1) and (2), considering that the force is positive (+) if it goes in the direction of the movement and negative (-) if it opposes the movement:

∑Fy= 0

N-Wp=0

N= Wp=170 N

∑Fx =0

F - Ff=0    F :force with which the player must push the disc

F - 0.0600*170 =0

F=10.2N