A player uses a hockey stick to push a puck at a constant velocity across the ice. The weight of the puck is 1.70 N. The coeffic

Question

3 years ago

9

A player uses a hockey stick to push a puck at a constant velocity across the ice. The weight of the puck is 1.70 N. The coeffic

ient of friction is 0.0600. With what force must the player push the puck so that his force just equals the frictional force?

Chemistry

2 answers:

Mekhanik [1.2K] · Answer 1 · 2022-02-28T01:53:54+03:00

Mekhanik [1.2K]3 years ago

7 0

Force = mass × acceleration

static force = coefficient of friction × normal force
1.70 = 0.0600 × normal force...
normal force = 1.70 ÷ 0.0600.. to obtain....

elena-14-01-66 [18.8K] · Answer 2 · 2022-02-28T03:48:29+03:00

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 = $\mu _f=0.0600$

Frictional force = $F_f=\mu _f\times N=0.0600\times 1.70 N=0.102 N$

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