/*! 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 2 A block of mass \(2 \cdot 00 \ma... [FREE SOLUTION] | 91Ó°ÊÓ

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A block of mass \(2 \cdot 00 \mathrm{~kg}\) moving at a speed of \(10^{\circ} 0 \mathrm{~m} / \mathrm{s}\) accelerates at \(3 \cdot 00 \mathrm{~m} / \mathrm{s}^{2}\) for \(5 \cdot 00 \mathrm{~s}\). Compute its final kinetic energy.

Short Answer

Expert verified
The final kinetic energy is 625.0 Joules.

Step by step solution

01

Understanding the Problem

We need to calculate the final kinetic energy of a block with a given mass moving at a certain initial speed and accelerating at a known rate over a specified time period. Since the kinetic energy formula involves velocity, we first need to determine the final velocity of the block.
02

Finding Final Velocity

Use the equation of motion to find the final velocity \( v_f \):\[ v_f = v_i + a imes t \]where \( v_i = 10.0 \text{ m/s} \) (initial velocity), \( a = 3.0 \text{ m/s}^2 \) (acceleration), and \( t = 5.0 \text{ s} \) (time).Substitute the values into the equation:\[ v_f = 10.0 + (3.0 \times 5.0) = 10.0 + 15.0 = 25.0 \text{ m/s} \]
03

Calculate Final Kinetic Energy

Use the kinetic energy formula:\[ KE = \frac{1}{2} m v_f^2 \]where \( m = 2.0 \text{ kg} \) (mass) and \( v_f = 25.0 \text{ m/s} \) (final velocity). Substitute these into the formula:\[ KE = \frac{1}{2} \times 2.0 \times (25.0)^2 \]\[ KE = 1.0 \times 625.0 = 625.0 \text{ J} \]So, the final kinetic energy of the block is 625.0 Joules.

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Key Concepts

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

Equations of Motion
Equations of motion play a crucial role in physics as they help predict the behavior of dynamic systems. These equations, derived from Newton's laws of motion, relate the physical quantities of velocity, acceleration, time, and displacement. The equation used in this exercise to solve for final velocity is:\[ v_f = v_i + a \times t \]This equation tells us that the final velocity (\( v_f \)) of an object is the sum of its initial velocity (\( v_i \)) and the product of acceleration (\( a \)) and time (\( t \)). Understanding how to manipulate these equations allows you to solve for unknown variables when given sufficient data.
  • Initial Velocity (\( v_i \)): The speed at which an object starts.
  • Acceleration (\( a \)): The rate of change of velocity. It can increase or decrease the speed of an object.
  • Time (\( t \)): How long the object has been accelerating.
By using this equation of motion, you can calculate how fast an object is moving after a certain period of time if you know the starting speed and how fast it is speeding up or slowing down.
Acceleration
Acceleration describes how quickly an object's velocity changes. It's not just about going faster; it can also mean slowing down or changing direction. In this exercise, the acceleration is given as \(3.0 \text{ m/s}^2\) which indicates a steady increase in velocity.Some important points about acceleration include:
  • Measured in meters per second squared (\(\text{ m/s}^2\)), reflecting how much the velocity changes each second.
  • Can be positive (speeding up) or negative (slowing down), often referred to as deceleration.
  • Directly impacts the final velocity and subsequently the kinetic energy of an object.
To understand this better, imagine you are pushing a block across the floor. If you push harder, the block speeds up more rapidly - this represents greater acceleration. In our problem, the block accelerates steadily, meaning every second, its speed increases by 3 meters per second. Recognizing the effects of acceleration helps predict future motion and is vital for designing systems that involve movement.
Final Velocity
Final velocity represents the speed of an object at a specific point after it has undergone acceleration. It is a key factor in determining other quantities such as kinetic energy.In the example provided, the final velocity is calculated using:\[ v_f = v_i + a \times t \]Given that the initial velocity \( v_i \) is \( 10.0 \text{ m/s} \), the acceleration \( a \) is \( 3.0 \text{ m/s}^2 \), and the time duration \( t \) for which the block accelerates is \( 5.0 \text{ s} \), the final velocity \( v_f \) is found to be \( 25.0 \text{ m/s} \).
  • Importance: Knowing the final velocity allows us to compute the final kinetic energy, a measure of the mechanical energy an object has due to its motion.
  • Application: Understanding final velocity can aid in designing safer vehicles, setting speed limits, or predicting the outcomes of collisions.
By capturing how fast an object is ultimately moving, final velocity provides insight into both the immediate effects of acceleration and the energy transitions within physical systems.

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Most popular questions from this chapter

A uniform chain of length \(L\) and mass \(M\) overhangs a horizontal table with its two third part on the table. The friction coefficient between the table and the chain is \(\mu\). Find the work done by the friction during the period the chain slips off the table.

The bob of a stationary pendulum is given a sharp hit to impart it a horizontal speed of \(\sqrt{3 g l}\). Find the angle rotated by the string before it becomes slack.

A block of mass \(5 \cdot 0 \mathrm{~kg}\) is suspended from the end of a vertical spring which is stretched by \(10 \mathrm{~cm}\) under the load of the block. The block is given a sharp impulse from below so that it acquires an upward speed of \(2 \cdot 0\) \(\mathrm{m} / \mathrm{s}\). How high will it rise ? Take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\)

A block of mass \(2 \cdot 0 \mathrm{~kg}\) is pushed down an inclined plane of inclination \(37^{\circ}\) with a force of \(20 \mathrm{~N}\) acting parallel to the incline. It is found that the block moves on the incline with an acceleration of \(10 \mathrm{~m} / \mathrm{s}^{2}\). If the block started from rest, find the work done (a) by the applied force in the first second, (b) by the weight of the block in the first second and (c) by the frictional force acting on the block in the first second. Take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\).

A small heavy block is attached to the lower end of a light rod of length \(l\) which can be rotated about its clamped upper end. What minimum horizontal velocity should the block be given so that it moves in a complete vertical circle?

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