∙4.60 A block of mass m1=21.9 kg is at rest on a plane inclined at θ=30.0^∘ above the horizonta

step 1 So here we have the system, the free -border diagram of the system. We can apply Newton's second

∙4.60 A block of mass m1=21.9 kg is at rest on a plane...

$\bullet 4.60$ A block of mass $m_{1}=21.9 \mathrm{~kg}$ is at rest on a plane inclined at $\theta=30.0^{\circ}$ above the horizontal. The block is connected via a rope and massless pulley system to another block of mass $m_{2}=25.1 \mathrm{~kg}$, as shown in the figure. The coefficients of static and kinetic friction between block 1 and the inclined plane are $\mu_{s}=0.109$ and $\mu_{k}=0.086$ respectively. If the blocks are released from rest, what is the displacement of block 2 in the vertical direction after 1.51 s? Use positive numbers for the upward direction and negative numbers for the downward direction.

Key Concepts

Dynamics of Connected Bodies

When two objects are connected by a rope over a pulley, their motions are interdependent. Understanding the constraints imposed by the rope and treating the pulley as massless (and frictionless, if specified) leads to the conclusion that both objects have related accelerations and tensions. This interrelationship is key to setting up and solving equations for their motion.

Newton's Second Law of Motion

This concept states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. It forms the basis for analyzing the dynamics of objects in motion, particularly when they are subjected to multiple forces along different directions.

Force Decomposition on an Inclined Plane

This involves breaking the gravitational force into components that are parallel and perpendicular to the surface of the incline. Understanding how to decompose forces allows for the proper calculation of the net force acting along the incline and is essential when analyzing motion on slopes.

Frictional Forces

Friction is the resistive force that opposes motion between two surfaces in contact. The concept includes both static friction, which acts when objects are at rest to prevent motion, and kinetic friction, which acts when objects are in motion. Correctly identifying which type of friction is relevant and incorporating its coefficient is vital for accurately predicting the behavior of objects under various conditions.

Kinematics Equations for Constant Acceleration

These equations describe the relationship between displacement, initial velocity, acceleration, and time for objects moving with constant acceleration. They provide a means to calculate how far an object travels over a given time interval, once the acceleration has been determined through dynamic analysis.

Transcript

00:01 So here we have the system, the free -border diagram of the system.

00:05 We can apply newton's second law in the x direction for the first mass.

00:11 This would be equaling to m sub 1 times the acceleration in the x direction.

00:17 We have then the tension minus m sub 1g sine of theta minus the kinetic frictional force equaling m sub 1 times the acceleration in the x direction.

00:32 And so the tension force t would be equaling to m sub 1a plus m sub 1g sine of theta plus the coefficient of kinetic friction multiplied by the normal force.

00:54 For the y direction, sum of forces in the y direction for the first mass, here we have then n minus m sub 1 g cosine of theta and the acceleration in the wide direction is 0 and so this is equaling 0 this will be equal to 0 and so the normal force n equaling m sub 1 g cosine of theta we can substitute this in so plug this into here and we have that the tension equals m sub 1.

01:34 A plus m sub 1g sine of theta plus the coefficient of kinetic friction m sub 1 g cosine of theta for the second mass we can apply newton's second law in the wide direction this would be equal to m sub 2 times the acceleration in the wide direction we have m sub 2 g minus the tension equaling m sub 2 times the acceleration in the y direction or simply a because this would be denoted as the acceleration of the system and so the tension t would be equaling to m sub 2 multiplied by then g minus a and so from this we can find the acceleration of the blocks and so we have essentially m m sub 2 multiplied by g minus m sub 2 multiplied by a, equaling m sub 1a plus m sub 1g sign of theta plus the coefficient of kinetic friction m sub 1g cosine of theta.

03:03 And solving for the acceleration a, this would be equal to then m sub 2g minus m sub 1.

03:12 G multiplied by sine of theta plus the coefficient of kinetic friction multiplied by cosine of theta.

03:24 And this all would be divided by the sum of the masses...

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