91Ó°ÊÓ

At a certain factory, \(300 \mathrm{~kg}\) crates are dropped vertically from a packing machine onto a conveyor belt moving at \(1.20 \mathrm{~m} / \mathrm{s}\) (Fig. \(8-62\) ). (A motor maintains the belt's constant speed.) The coefficient of kinetic friction between the belt and each crate is \(0.400\). After a short time, slipping between the belt and the crate ceases, and the crate then moves along with the belt. For the period of time during which the crate is being brought to rest relative to the belt, calculate, for a coordinate system at rest in the factory, (a) the kinetic energy supplied to the crate, (b) the magnitude of the kinetic frictional force acting on the crate, and (c) the energy supplied by the motor. (d) Explain why answers (a) and (c) differ.

Step by step solution

01

Understanding the Problem

We are dealing with a situation where crates are dropped onto a moving conveyor belt. We need to calculate the kinetic energy supplied to the crate (part a), the kinetic friction force acting on the crate (part b), and the energy supplied by the motor (part c). Finally, we explain why there is a difference in energies (part d).

02

Calculate the Kinetic Energy Supplied (Part A)

Initially, the crate is at rest relative to the moving belt, so its relative velocity is the same as the belt's velocity, which is \(1.20 \mathrm{~m/s}\). The kinetic energy supplied to the crate will be the change in its kinetic energy as it reaches the belt's speed. Since the mass of the crate \(m = 300 \mathrm{~kg}\) and its velocity \(v = 1.20 \mathrm{~m/s}\), then the kinetic energy is: \[KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 300 \times (1.20)^2 = 216 \mathrm{~J}\]

03

Determine the Kinetic Friction Force (Part B)

The force of kinetic friction \(f_k\) can be determined by \(f_k = \mu_k m g\), where \(\mu_k = 0.400\) is the coefficient of kinetic friction and \(g = 9.81 \mathrm{~m/s^2}\) is the acceleration due to gravity. Thus, \[f_k = 0.400 \times 300 \times 9.81 = 1177.2 \mathrm{~N}\]

04

Calculate the Energy Supplied by the Motor (Part C)

The work done by the kinetic friction force to stop the relative motion between the crate and the belt equals the energy supplied by the motor to maintain the speed of the belt throughout the crate's adjustment period. Calculate the energy by multiplying the friction force \(f_k\) by the distance \(d\) the crate travels while slowing down relative to the belt. We don't have \(d\) directly, but can understand that this work is equal to the kinetic energy calculated, so it is also \(216 \mathrm{~J}\).

05

Explaining the Difference (Part D)

The difference between the kinetic energy supplied to the crate (216 J) and the energy supplied by the motor (also 216 J) in this ideal scenario is zero due to assumptions of perfect energy transfer and no external losses like air resistance. In more realistic settings, motor work may exceed energy supplied to the crate due to inefficiencies.

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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 possessed by a body due to its motion. It's a fundamental concept in physics that helps us understand how objects move.
To calculate the kinetic energy (\( KE \)) of an object, we use the formula:

• \[ KE = \frac{1}{2} mv^2 \]where \( m \) is the mass and \( v \) is the velocity of the object.

In our problem, a crate is initially at rest relative to a conveyor belt moving at \( 1.20 \, \text{m/s} \). The energy needed to bring the crate to this speed is its kinetic energy.
The mass of the crate is \( 300 \, \text{kg} \) and the change in speed leads to a new kinetic energy of \( 216 \, \text{J} \). This calculated energy helps us understand how much work was necessary to get the crate moving at that speed.

Kinetic Friction

Kinetic friction is the resistive force that acts when two surfaces slide over each other. It's essential in many physics problems as it determines how quickly an object will slow or speed up when in contact with another surface.
The force of kinetic friction (\( f_k \)) is given by:

• \[ f_k = \mu_k m g \]where \( \mu_k \) is the coefficient of kinetic friction, \( m \) is the mass, and \( g \) is the gravitational acceleration (approx. \( 9.81 \, \text{m/s}^2 \)).

In our scenario, the force of kinetic friction acting on the crate is calculated to be \( 1177.2 \, \text{N} \). This force opposes the motion of the crate relative to the belt, causing it to eventually match the speed of the belt.

Energy Conservation

Energy conservation is a principle indicating that the total energy in an isolated system remains constant. It beautifully ties into understanding how energy is transferred and transformed.
In our example, as the crate interacts with the conveyor belt, energy is neither created nor destroyed; it is merely converted from one form to another.

• The kinetic energy gained by the crate as it speeds up equals the work done by the motor to maintain the conveyor's speed.
• This means the energy supplied by the motor effectively matches the kinetic energy required to accelerate the crate to the belt's speed, calculated to be \( 216 \, \text{J} \).
• Ideally, without external losses, such as air resistance, the energies match perfectly, illustrating efficient energy conservation.

Understanding energy conservation allows us to predict and calculate changes in systems efficiently and effectively.