A wooden block with mass 0.26{\rm kg} rests on a horizontal table, connected to a string that hangs vertically over a friction-less pulley on the table's edge. From the other end of the string hangs a 0.20{\rm kg} mass.

What minimum coefficient of static friction \mu _{{\rm s}} between the block and table will keep the system at rest?

Find the block's acceleration if \mu _k =0.17.

To find the minimum coefficient of static friction that will keep the system at rest, we need to set up the equations of motion for both masses.

1. For the wooden block:
The forces acting on the wooden block are the tension force T in the string and the friction force f. The weight of the block is balanced by the normal force from the table.
The equation of motion for the block in the horizontal direction is:
f = μ_s * N
where μ_s is the coefficient of static friction, and N is the normal force.
The normal force N = mass * gravity.
The equation of motion for the block in the vertical direction is:
N - mg = 0
where m is the mass of the wooden block and g is the acceleration due to gravity.
Since the system is at rest, the tension force T is equal to the weight of the hanging mass (T = mg).

2. For the hanging mass:
The force acting on the hanging mass is its weight mg.
The equation of motion for the hanging mass is:
T - mg = 0

Now, we can solve for the minimum coefficient of static friction μ_s:

For the wooden block:
N - mg = 0
N = mg

f = μ_s * N
f = μ_s * mg

For the hanging mass:
T - mg = 0
T = mg

Since T = f, we can equate the two expressions:

μ_s * mg = mg

Dividing both sides by mg:

μ_s = 1

So the minimum coefficient of static friction μ_s that will keep the system at rest is 1.

To find the block's acceleration when the coefficient of kinetic friction μ_k = 0.17, we can use the same equations of motion, but now the friction force is given by:

f = μ_k * N

Substituting this expression into the equation for the block's motion in the horizontal direction:

μ_k * N = m * a

Solving for acceleration:

a = (μ_k * N) / m

Using the previous equation N = mg, we have:

a = (μ_k * mg) / m

Plugging in the values:

a = (0.17 * 0.26 * 9.8) / 0.26

Simplifying the expression:

a = 0.17 * 9.8

a = 1.666 m/s²

So, the block's acceleration is 1.666 m/s² when the coefficient of kinetic friction μ_k = 0.17.

To find the minimum coefficient of static friction, we need to analyze the forces acting on the system. The forces involved are the gravitational force (weight) and the tension in the string.

Let's start by finding the tension in the string. Since the system is at rest, the upward tension force must exactly balance the downward gravitational force. Now, we can find the tension in the string using Newton's second law.

The gravitational force on the 0.20 kg mass is given by:
F_gravity = m * g = (0.20 kg) * (9.8 m/s^2)
F_gravity = 1.96 N

Since the system is at rest, the tension in the string must also be 1.96 N. This tension force is pulling the wooden block towards the left.

Now, let's analyze the forces on the wooden block. The forces acting on the block are its weight (mg) and the tension force (T) pulling it towards the left.

The net force acting on the block is given by:
F_net = F_tension - F_friction

Since the wooden block is at rest, the minimum coefficient of static friction can be found when the net force is zero. Therefore, we set F_net = 0 and solve for the minimum coefficient of static friction.

F_net = F_tension - F_friction
0 = T - F_friction

The maximum static friction force can be expressed as:
F_friction = μ_s * N

where μ_s is the coefficient of static friction and N is the normal force on the block. In this case, the normal force is equal to the weight of the block, N = mg.

Substituting these values into the equation:
0 = T - μ_s * mg

We know that T = 1.96 N, m = 0.26 kg, and g = 9.8 m/s^2. Substituting these values, we can solve for the minimum coefficient of static friction, μ_s:

0 = 1.96 N - μ_s * (0.26 kg)(9.8 m/s^2)

Solving for μ_s, we get:
μ_s = 1.96 N / (0.26 kg * 9.8 m/s^2)

Now, let's calculate the block's acceleration with a coefficient of kinetic friction (μ_k) of 0.17. The frictional force (F_friction) acting on the block is given by:

F_friction = μ_k * N

Substituting F_friction = μ_k * mg into the equation for net force:
F_net = T - F_friction
ma = T - μ_k * mg

We know that T = 1.96 N, m = 0.26 kg, g = 9.8 m/s^2, and μ_k = 0.17. Substituting these values, we can solve for the acceleration (a):

ma = 1.96 N - (0.17)(0.26 kg)(9.8 m/s^2)

Now, solve for a:
a = (1.96 N - (0.17)(0.26 kg)(9.8 m/s^2)) / 0.26 kg

By evaluating this expression, we can find the block's acceleration.