1(a).Given is a box (Object 1 with a mass of 𝑚 ) on a table connected to a1= 1 𝑘𝑔box is a rope which has no mass. The rope passess over a frictionless, masslesspulley and is fixed to another box (Object 2 with a mass 𝑚 ). The pulley has2= 2 𝑘𝑔a fixed position and cannot move left, right up or down. Object 1 is being accelerated5. 41 𝑚/𝑠2. What is the friction coefficient of Object 1 and table top? Assume𝑔 = 9. 8 𝑚/𝑠2 and let us agree that moving to the left is the positive direction ofobject 1.(b) Object 𝑀 on inclined at an angle connected to a rope with no1= 20 𝑘𝑔 α = 20°mass. The rope passes over frictionless, massless. Pulley is fixed to the ceiling andcannot move. Pulley is hanging on a rope through which a second box is attached.Pulley moves up if object 1 goes down the incline. static and kinetic coefficientµ. Find the so object 1 is not moving.

1(a) To find the friction coefficient of Object 1 and the table top, we can use Newton's second law of motion.

First, let's draw a free body diagram for Object 1. We have the force of gravity acting downwards with a magnitude of mg, where m is the mass of Object 1 and g is the acceleration due to gravity (9.8 m/s^2). We also have the tension in the rope pulling Object 1 upwards, and the friction force opposing the motion of Object 1.

Since Object 1 is being accelerated, we can write the equation of motion as:

ma = T - mg - f

where a is the acceleration of Object 1, T is the tension in the rope, and f is the friction force.

We are given that the acceleration of Object 1 is 5.41 m/s^2. Plugging in the given values, we have:

m(5.41) = T - mg - f

Simplifying the equation, we get:

T - mg - f = 5.41m

Now, let's consider Object 2. Since the pulley is frictionless and massless, the tension in the rope on the other side of the pulley is also T. Therefore, the force acting on Object 2 is T in the upward direction.

Using Newton's second law for Object 2, we have:

ma = T

where a is the acceleration of Object 2. Since Object 2 is not accelerating, we have:

0 = T

This means that the tension in the rope is zero.

Now, let's go back to the equation for Object 1 and substitute T = 0:

0 - mg - f = 5.41m

Simplifying further, we get:

-f = 5.41m - mg

Since the friction force is given by f = μN, where μ is the friction coefficient and N is the normal force, we can rewrite the equation as:

-μN = 5.41m - mg

The normal force N is equal to mg, so we have:

-μmg = 5.41m - mg

Simplifying, we get:

(5.41 - 1)mg = μmg

Dividing both sides by mg, we get:

5.41 - 1 = μ

Therefore, the friction coefficient of Object 1 and the table top is 4.41.

(b) To find the static and kinetic coefficients of friction, we need to consider the forces acting on Object M on the inclined plane.

Let's draw a free body diagram for Object M. We have the force of gravity acting downwards with a magnitude of mg, where m is the mass of Object M and g is the acceleration due to gravity (9.8 m/s^2). We also have the normal force N acting perpendicular to the inclined plane, and the friction force opposing the motion of Object M.

Since Object M is not moving, the force of gravity can be resolved into two components: one parallel to the inclined plane and one perpendicular to the inclined plane. The component parallel to the inclined plane is mgsin(α), where α is the angle of inclination. The component perpendicular to the inclined plane is mgcos(α).

The friction force is given by f = μN, where μ is the friction coefficient and N is the normal force. In this case, the friction force is opposing the motion of Object M, so it acts in the opposite direction of the component parallel to the inclined plane.

Using Newton's second law for Object M, we have:

ma = mg*sin(α) - f

Since Object M is not moving, the acceleration is zero. Therefore, we have:

0 = mg*sin(α) - f

Substituting f = μN, we get:

0 = mg*sin(α) - μN

The normal force N is equal to mg*cos(α), so we have:

0 = mgsin(α) - μmgcos(α)

Simplifying further, we get:

μmgcos(α) = mgsin(α)

Dividing both sides by mg*cos(α), we get:

μ = tan(α)

Therefore, the static and kinetic coefficients of friction are both equal to tan(α), where α is the angle of inclination.