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An ice skater applies a horizontal force to a 150 N block on frictionless, level ice, causing the block to accelerate uniformly at to the right. After the skater
stops pushing the block, it slides onto a region of ice that is covered with a thin layer of sand. The coefficient of kinetic friction between the block and the
sand-covered ice is 0.36.
Calculate the magnitude of the force of friction acting on the block as it slides over the sand-covered ice

Answer :

elcharly64

Answer:

The friction force has a magnitude of 54 N and points to the left

Explanation:

Friction Force

When an object is moving and encounters friction in the air or rough surfaces, it loses acceleration because the friction force opposes motion.

The friction force when an object is moving on a horizontal surface is calculated by:

[tex]Fr=\mu N[/tex]

Where \mu is the coefficient of static or kinetics friction and N is the normal force.

If no forces other then the weight and the normal are acting upon the y-direction, then the weight and the normal are equal in magnitude:

N = W

Thus, the friction force is:

[tex]Fr=\mu W[/tex]

The ice skater accelerates a block to the right on frictionless horizontal ice and then stops pushing it. The block continues to move at a constant speed when it finds a sand-covered surface with a kinetic coefficient of friction of 0.36.

The friction force is:

Fr=0.36*150

Fr = 54 N

The friction force has a magnitude of 54 N and points to the left

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