Question

\(\bullet 4.60\) 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.

Short answer

Please provide the necessary information, such as the masses of the blocks, the coefficient of static and kinetic friction, and the angle of the inclined plane.

Step-by-step solution

Identify the forces acting on the blocks

We must consider the gravitational force acting on each mass, the tension in the rope, and the friction force acting on block 1 due to the inclined plane. And also take note that the angle of incline is \(30.0^{\circ}\).

Calculate the gravitational force acting on each block

The gravitational force for each block is equal to the mass multiplied by the acceleration due to gravity (\(g = 9.81 m/s^2\)). For block 1, \(F_{g1} = m_{1}g\); for block 2, \(F_{g2} = m_{2}g\).

Calculate the gravitational force components for block 1

In order to find the frictional force acting on block 1, we need to find the components of the gravitational force parallel and perpendicular to the inclined plane. The parallel component \(F_{parallel} = F_{g1}\sin{\theta}\) and the perpendicular component \(F_{perpendicular} = F_{g1}\cos{\theta}\).

Calculate the static friction force and check if the blocks move

The maximum static frictional force that can act on block 1 is given by \(F_s = \mu_s F_{perpendicular}\). If the gravitational force component parallel to the incline is greater than the maximum static friction force, the blocks will move. Check whether \(F_{parallel} > F_s\); if true, proceed to Step 5.

Calculate the kinetic friction force

Since the blocks move, we need to calculate the kinetic friction force that will act on block 1. The kinetic friction force is given by \(F_k = \mu_k F_{perpendicular}\).

Calculate the net force acting on both blocks

Now that we have all forces acting on block 1, we can find the net force acting on both blocks. The net force on block 1 is \(F_{net1} = F_{parallel} - F_k\), and the net force on block 2 is \(F_{net2} = F_{g2} - T\). As the blocks move together, we can write the equation for the total net force as \(F_{total} = F_{net1} + F_{net2}\).

Calculate the acceleration of the system

Using Newton's second law equation \(F = ma\), we can find the acceleration of the system with the total net force. The total mass of the system is \(m_{total} = m_{1}+m_{2}\). Substitute and solve for acceleration: \(a = \frac{F_{total}}{m_{total}}\).

Determine the displacement of block 2 using the kinematic equation

As block 2 is initially at rest, we can use the kinematic equation \(d = v_0t + \frac{1}{2}at^2\) to find the displacement of block 2 in the vertical direction. Block 2's initial velocity \(v_0 = 0\) and the time \(t = 1.51 s\). Substitute the values and the calculated acceleration to find the displacement of block 2. Remember to use positive numbers for the upward direction and negative numbers for the downward direction.

Key concepts

Static and Kinetic Friction

Static and kinetic friction are two essential forces to understand in inclined plane motion. **Static friction (\(F_s\))** prevents an object from moving when it's at rest. Its strength depends on the normal force and the coefficient of static friction \(\mu_s\). In this problem, the normal force for block 1 on the incline is calculated as the perpendicular component of gravity.
The formula for maximum static friction is:

• \(F_s = \mu_s F_{perpendicular}\)

Block 1 only starts moving when the force parallel to the incline exceeds \(F_s\). If the blocks move, static friction gives way to kinetic friction.
**Kinetic friction (\(F_k\))** acts on moving objects, opposing the motion. It's usually weaker than static friction due to \(\mu_k < \mu_s\), leading to easier movement once the object starts sliding. The kinetic friction force is calculated similarly:

• \(F_k = \mu_k F_{perpendicular}\)

Newton's Laws of Motion

Newton's Laws of Motion form the basis for understanding motion on inclined planes. **The First Law** indicates that an object remains at rest or in motion unless acted upon by a unbalanced force. In our problem, this relates to overcoming static friction.
**The Second Law**, \(F = ma\), explains how forces cause acceleration. It's crucial here when calculating the acceleration of the entire system, as it combines all acting forces:

• From friction
• From gravity
• From tension in the rope

We find acceleration by dividing the total net force acting on both blocks by their combined mass. This equation helps in determining the eventual movement and displacement of block 2.
**The Third Law**, however, is less directly applicable to this specific setup but suggests that for every action there is an equal and opposite reaction, which helps us think about the forces between the two blocks.

Kinematic Equations

Kinematic equations describe the motion of objects without considering the forces causing the motion. They're handy when initial velocity, time, and acceleration are known—a classic case with inclined plane problems.
For block 2's displacement, we utilize the equation:

• \(d = v_0t + \frac{1}{2}at^2\)

Here, block 2 starts from rest, making \(v_0 = 0\). The acceleration \(a\) is known from the forces calculated using Newton's laws, and time \(t\) is given as 1.51 s.
This equation tells us how far block 2 moves vertically in the given time, driven by the net forces acting on the system. It's used at the final step of the solution after all forces and accelerations are determined.

Gravitational Force Components

When dealing with inclined planes, gravity must be split into two components to understand its effect clearly.
**Parallel Component (\(F_{parallel}\))**: This force acts along the plane and tends to pull the object down the incline. Calculated using:

• \(F_{parallel} = F_{g1} \sin(\theta)\)

It plays a key role in determining whether block 1 starts sliding or remains stationary.
**Perpendicular Component (\(F_{perpendicular}\))**: This acts perpendicular to the plane, influencing the normal force. It's calculated as:

• \(F_{perpendicular} = F_{g1} \cos(\theta)\)

The normal force impacts the frictional forces working against motion. By decomposing gravitational force, it makes analyzing motion on an incline straightforward.