/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 A \(2 \mathrm{~kg}\) block is mo... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(2 \mathrm{~kg}\) block is moving at \(5 \mathrm{~m} / \mathrm{s}\) along a frictionless table and collides with a second \(2 \mathrm{~kg}\) block that is initially at rest. After the collision, the two blocks stick together and then slide up a \(45^{\circ}\) frictionless inclined plane, as shown in Figure \(8.41 .\) Calculate the maximum distance \(L\) that the two blocks travel up the incline.

Short Answer

Expert verified
The maximum distance the blocks travel up the incline is approximately 0.45 m.

Step by step solution

01

Compute Initial Momentum

Since the first block is the only one moving initially, we calculate the initial momentum of the system using the formula: \( p = m_1 \cdot v_1 \).Substitute the given values:\[ p = 2 \mathrm{~kg} \times 5 \mathrm{~m/s} = 10 \mathrm{~kg \, m/s}. \]
02

Compute Velocity After Collision

As the blocks stick together, they move as a single entity post-collision. Applying conservation of momentum to find the shared velocity \( v_f \):\[ m_1 \cdot v_1 = (m_1 + m_2) \cdot v_f. \]Substitute the known values:\[ 10 \mathrm{~kg \, m/s} = (2 \mathrm{~kg} + 2 \mathrm{~kg}) \cdot v_f, \]which simplifies to:\[ v_f = \frac{10}{4} \mathrm{~m/s} = 2.5 \mathrm{~m/s}. \]
03

Find Initial Kinetic Energy After Collision

Calculate the kinetic energy of the blocks post-collision:\[ K = \frac{1}{2} (m_1 + m_2) v_f^2. \]Substitute the values:\[ K = \frac{1}{2} (4 \mathrm{~kg}) (2.5 \mathrm{~m/s})^2, \]which yields:\[ K = \frac{1}{2} \times 4 \times 6.25 = 12.5 \mathrm{~J}. \]
04

Apply Energy Conservation to Calculate Maximum Height

As the blocks slide up the incline, their kinetic energy is converted to potential energy (\( U = mgh \)). By conservation of mechanical energy:\[ K = U \Rightarrow mgh = 12.5 \mathrm{~J}. \]Rearranging, we get:\[ h = \frac{12.5}{4 \cdot 9.8} \mathrm{~m}. \]This simplifies to:\[ h = 0.318 \mathrm{~m}. \]
05

Calculate Maximum Distance Up the Incline

Use trigonometry to relate the height \( h \) to the distance \( L \) up the incline. For a \(45^{\circ}\) incline:\[ \sin(45^{\circ}) = \frac{h}{L}. \]Rearranging gives:\[ L = \frac{0.318}{\sin(45^{\circ})} \mathrm{~m}, \]which simplifies to:\[ L = \frac{0.318}{0.7071} \approx 0.45 \mathrm{~m}. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It depends on both the mass of the object and the square of its velocity. The formula to calculate kinetic energy (K) is: \[ K = \frac{1}{2} m v^2 \]where:
  • \(m\) is the mass of the object
  • \(v\) is the velocity of the object
Kinetic energy increases with both the mass and velocity. This means a heavier object will have more kinetic energy if it moves at the same speed as a lighter object. Similarly, an object that moves faster will have more kinetic energy if it has the same mass.
In our exercise, before the collision, only the first block is moving. This motion gives it kinetic energy, calculated using its mass and speed. After colliding, the energy is shared between both blocks as they move together.
Potential Energy
Potential energy is stored energy due to an object’s position or state. When thinking about potential energy in a gravitational field:
  • Higher elevation means higher potential energy.
  • It is given by the formula: \( U = mgh \), where:
  • \(m\) is the mass of the object
  • \(g\) is the acceleration due to gravity
  • \(h\) is the height above the reference point
The potential energy increases as the blocks are pushed up the incline. When the kinetic energy from the motion is completely converted to potential energy, it reaches its maximum value, and the blocks momentarily stop, at the highest point they achieve on the plane.
This conversion from moving (kinetic) to height (potential) energy is vital to quantifying how far the blocks move up the incline.
Frictionless Inclined Plane
An inclined plane is a flat surface tilted at an angle, like a ramp. It helps in moving heavy objects with less energy. A frictionless inclined plane means there's no resistance slowing the objects.
When no friction is present:
  • Objects slide without losing energy to heat.
  • The movement depends solely on the force applied and gravitational pull.
For our problem, the frictionless plane means the blocks glide up with no energy lost. The only forces at play are the block's kinetic energy and gravity pulling them down. All kinetic energy turns to potential energy as blocks rise, making calculations simpler.
The frictionless nature ensures that the mechanical energy conservation law holds true without any loss due to thermal energy or other resistive forces.
Mechanical Energy Conservation
Mechanical energy conservation is a key principle in physics, stating the total mechanical energy stays constant if only conservative forces (like gravity) act on the system.
In other words:
  • The sum of kinetic energy and potential energy before an event equals the sum afterwards.
  • Formally: \( K_i + U_i = K_f + U_f \), where \(K\) and \(U\) are the kinetic and potential energies initially (\(i\)) and finally (\(f\)).
In the problem given, after the collision, the kinetic energy of the blocks is transformed into potential energy as they ascend the incline. By using the mechanical energy conservation law, we calculate how much kinetic energy is converted and thus, find how high and far the blocks can travel proportionately up the incline.
This ensures every bit of initial mechanical energy quantifies physical changes, like speed reductions or altitude gains, proving efficiencies absent of external forces such as friction.

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Most popular questions from this chapter

To protect their young in the nest, peregrine falcons will fly into birds of prey (such as ravens) at high speed. In one such episode, a \(600 \mathrm{~g}\) falcon flying at \(20.0 \mathrm{~m} / \mathrm{s}\) flew into a \(1.5 \mathrm{~kg}\) raven flying at \(9.0 \mathrm{~m} / \mathrm{s}\). The falcon hit the raven at right angles to its original path and bounced back with a speed of \(5.0 \mathrm{~m} / \mathrm{s} .\) By what angle did the falcon change the raven's direction of motion?

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