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Chapter 4: Problem 60

\(\bullet 4.60\) 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

Expert verified
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

01

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 \(30.0^{\circ}\).
02

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, \(F_{g2} = m_{2}g\).
03

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 \(F_{perpendicular} = F_{g1}\cos{\theta}\).
04

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.
05

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 \(F_k = \mu_k F_{perpendicular}\).
06

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 \(F_{total} = F_{net1} + F_{net2}\).
07

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: \(a = \frac{F_{total}}{m_{total}}\).
08

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.

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Most popular questions from this chapter

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