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the diagram below represents a 4.0-newton force applied to a 0.200-kilogram copper block sliding to the right on a horizontal steel table

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Question is not complete, so i have attached it.

Answer:

A) W = 1.962 N

B) F_f = 0.706 N

C) F_net = 3.294 N

Explanation:

A) The block is 0.2 kg.

Formula for weight is;

W = mg

W = 0.2 × 9.81

W = 1.962 N

B) Magnitude of the force acting on the moving block is given by;

F_f = μmg

Where μ is coefficient of kinetic friction

Coefficient of kinetic friction for copper(mild steel) from online sources is 0.36

Thus;

F_f = 0.36 × 1.962

F_f = 0.706 N

C) Magnitude of net force is;

F_net = F_applied - F_f

F_net = 4 - 0.706

F_net = 3.294 N

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Lanuel

1. The weight of the copper block is equal to 1.96 Newton.

2. The magnitude of the force acting on the moving copper block is 0.7056 Newton.

3. The magnitude of the net force acting on the moving copper block is 3.2944 Newton.

Given the following data:

  • Force = 4 Newton
  • Mass of copper block = 0.200 Kg

We know that the acceleration due to gravity (g) of an object on planet Earth is equal to 9.8 [tex]m/s^2[/tex].

We know that the coefficient of kinetic friction for copper is equal to 0.36.

1. To determine the weight of the copper block:

[tex]Weight = mg\\\\Weight = 0.200 \times 9.8[/tex]

Weight = 1.96 Newton

2. To determine the magnitude of the force acting on the moving copper block:

Mathematically, the force of kinetic friction is given by the formula;

[tex]F_k = umg[/tex]

Where;

  • u is the coefficient of kinetic friction.
  • g is the acceleration due to gravity.
  • m is the mass of an object.

Substituting the parameters into the formula, we have;

[tex]F_k = 0.36 \times 0.200 \times 9.8\\\\F_k = 0.7056 \;Newton[/tex]

3. To determine the magnitude of the net force acting on the moving copper block:

[tex]F_{net} = F - F_k\\\\F_{net} = 4-0.7056\\\\F_{net} = 3.2944\;Newton[/tex]

Read more: https://brainly.com/question/13754413

Ver imagen Lanuel