A 1.00 kg box, initially at rest, is pulled across a table with a force of 3.00 N. (a) Will the box move if the coefficient of friction is 0.45? (b) What is the coefficient of friction if the box just begins to move? (c) What is the acceleration of the box if the coefficient of friction is 0.20?

To answer these questions, we need to use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration (F = ma). Additionally, we need to consider the force of friction, which opposes the motion of the box.

(a) Will the box move if the coefficient of friction is 0.45?
To determine if the box will move, we need to compare the force of friction with the applied force. The force of friction (Ff) can be calculated using the formula:

Ff = μ * Fn

where μ is the coefficient of friction and Fn is the normal force.

Since the box is initially at rest, the normal force is equal to the weight of the box, which can be calculated as:

Fn = mg

where m is the mass of the box and g is the acceleration due to gravity.

The force of friction is then:

Ff = μ * m * g

In this case, m = 1.00 kg, g = 9.8 m/s², and μ = 0.45.

Now, let's calculate the force of friction:

Ff = 0.45 * 1.00 kg * 9.8 m/s²

Ff ≈ 4.41 N

The force of friction (4.41 N) is greater than the applied force (3.00 N), so the box will not move.

(b) What is the coefficient of friction if the box just begins to move?
To determine the coefficient of friction (μ) when the box just begins to move, we need to find the threshold value at which the force of friction is equal to the applied force.

In this case, the box begins to move, so the force of friction can be expressed as:

Ff = μ * Fn

By substituting the normal force with the weight (Fn = mg), we have:

Ff = μ * m * g

Since the box is just starting to move, the force of friction is equal to the applied force:

Ff = Fa

where Fa is the applied force.

In this case, m = 1.00 kg, g = 9.8 m/s², and Fa = 3.00 N.

Now, let's solve for the coefficient of friction:

Fa = μ * m * g

3.00 N = μ * 1.00 kg * 9.8 m/s²

μ ≈ 0.31

So, the coefficient of friction is approximately 0.31 when the box just begins to move.

(c) What is the acceleration of the box if the coefficient of friction is 0.20?
To determine the acceleration of the box, we can use the formula:

ΣF = ma

where ΣF is the net force acting on the box.

In this case, the net force is equal to the applied force minus the force of friction:

ΣF = Fa - Ff

Let's substitute the values:

Fa = 3.00 N (given)

Ff = μ * Fn = μ * m * g = 0.20 * 1.00 kg * 9.8 m/s² = 1.96 N

ΣF = 3.00 N - 1.96 N = 1.04 N

Now, we can find the acceleration by dividing the net force by the mass:

a = ΣF / m

a = 1.04 N / 1.00 kg

a = 1.04 m/s²

Therefore, if the coefficient of friction is 0.20, the box will experience an acceleration of 1.04 m/s².