Question

A system with a mass of \(5 \mathrm{~kg}\), initially moving horizontally with a velocity of \(40 \mathrm{~m} / \mathrm{s}\), experiences a constant horizontal deceleration of \(2 \mathrm{~m} / \mathrm{s}^{2}\) due to the action of a resultant force. As a result, the system comes to rest. Determine the length of time, in s, the force is applied and the amount of energy transfer by work, in \(\mathrm{kJ}\).

Short answer

The force is applied for 20 seconds, and the work done is 4 kJ.

Step-by-step solution

- Understand the problem

A system with a mass of 5 kg is moving with an initial velocity of 40 m/s and experiences a constant deceleration of 2 m/s². The goal is to find the time it takes for the system to come to rest and the work done during this process.

- Use the equation of motion to find time

Use the kinematic equation: \[ v = u + at \] where: - \( v \) is the final velocity (0 m/s, since the system comes to rest) - \( u \) is the initial velocity (40 m/s) - \( a \) is the acceleration (-2 m/s², since it's a deceleration) Rearrange the equation to solve for time \( t \): \[ 0 = 40 - 2t \] \[ t = \frac{40}{2} = 20 \] So the time to come to rest is 20 seconds.

- Use Work-Energy Theorem to find work done

The work done (W) on the system is equal to the change in kinetic energy. The kinetic energy (KE) is given by: \[ \text{KE} = \frac{1}{2}mv^2 \] - Initial kinetic energy (KE_initial) when velocity is 40 m/s: \[ \text{KE_initial} = \frac{1}{2} \times 5 \times 40^2 = 4000 \text{ J} \] - Final kinetic energy (KE_final) when velocity is 0 m/s: \[ \text{KE_final} = \frac{1}{2} \times 5 \times 0^2 = 0 \text{ J} \] Work done (W) is the change in kinetic energy: \[ W = \text{KE_initial} - \text{KE_final} = 4000 - 0 = 4000 \text{ J} \] Convert the work from Joules to kilojoules: \[ 4000 \text{ J} = 4 \text{ kJ} \]

Key concepts

kinematic equations

Kinematic equations are fundamental to understanding motion in physics. They describe the behavior of a moving object under the influence of acceleration. There are four main kinematic equations that relate displacement, initial and final velocity, acceleration, and time. For this problem, we use the equation: oindent\[ v = u + at \] Here,

• \( v \) stands for the final velocity (0 m/s, as the system comes to rest)
• \( u \) represents the initial velocity (40 m/s)
• \( a \) is the constant acceleration (-2 m/s², since it is a deceleration)

Given these values, we rearrange to find a time, yielding: \[ t = \frac{v-u}{a} \] Plugging in our numbers, we get: \[ 0 = 40 - 2t \] Rearranging gives us: \[ t = \frac{40}{2} = 20 \] Thus, the system takes 20 seconds to come to rest. This illustrates how kinematic equations provide valuable insights into motion mechanics.

work-energy theorem

The work-energy theorem is another essential concept in physics. This principle states that the work done on an object is equal to the change in its kinetic energy. In mathematical terms, it is: \[ W = \text{KE}_\text{initial} - \text{KE}_\text{final} \] In the given problem, the initial and final kinetic energies are found using the formula for kinetic energy: \[ \text{KE} = \frac{1}{2}mv^2 \] With mass \( m = 5 \) kg and initial velocity \( u = 40 \) m/s, we calculate: \[ \text{KE}_\text{initial} = \frac{1}{2} \times 5 \times 40^2 = 4000 \text{ J} \] Since the system comes to rest, its final kinetic energy is: \[ \text{KE}_\text{final} = \frac{1}{2} \times 5 \times 0^2 = 0 \text{ J} \] The work done (W) on the system is thus: \[ W = 4000 - 0 = 4000 \text{ J} \] Converted to kilojoules: \[ W = 4 \text{ kJ} \] This theorem simplifies calculations by directly linking the kinetic energy change to the work applied, removing the need for intermediate steps.

deceleration

Deceleration is simply negative acceleration, indicating a decrease in an object's velocity. In the given problem, the system experiences a constant deceleration of \(2 \text{ m/s}^2\). The kinematic equation used, \[ v = u + at \] with \( a \) being negative, captures this deceleration. This leads to: \[ 0 = 40 - 2t \] Solving for time (\ t), we find it takes 20 seconds for the velocity to drop to zero. Deceleration is a critical concept when understanding how forces work to slow down or stop an object in motion. In this case, a resultant force applied over time leads to the system coming to a stop. The consistency of this deceleration is key to simplifying our calculations and understanding the behavior of the system.

kinetic energy

Kinetic energy (KE) is the energy possessed by an object due to its motion. The kinetic energy of an object is given by: \[ \text{KE} = \frac{1}{2}mv^2 \] where \( m \) is mass and \( v \) is velocity. For the system in the problem, with mass \( m = 5 \) kg and initial velocity \( v = 40 \) m/s, the initial kinetic energy is: \[ \text{KE}_\text{initial} = \frac{1}{2} \times 5 \times 40^2 = 4000 \text{ J} \] When the velocity decreases to 0 m/s, the kinetic energy becomes: \[ \text{KE}_\text{final} = \frac{1}{2} \times 5 \times 0^2 = 0 \text{ J} \] This change in kinetic energy (from 4000 J to 0 J) represents the work done (4 kJ) to bring the system to rest. Kinetic energy calculations allow us to understand how much work is needed to change the motion state of a system. In applications such as vehicles braking, sports, and machinery, knowing kinetic energy offers insights into the energy dynamics at play.