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Debora [2.8K]
3 years ago
14

A 3-kg block rests on top of a 2-kg block supported by, but not attached to, a spring of constant 40 N/m. The upper block is sud

denly removed. Determine (a) the maximum speed reached by the 2-kg block, (b) the maximum height reached by the 2-kg block.

Engineering
2 answers:
Inessa05 [86]3 years ago
8 0
<h2></h2><h2>Answer:</h2><h2></h2>

(a) 5.59m/s

(b) 1.56m

<h2>Explanation:</h2>

<em></em>

<em>Analysis 1:</em> When the 3-kg block rests on the 2-kg block, their total weight, W is given as;

W = M x g       -------------------(i)

Where;

M = total mass = 3kg + 2 kg = 5kg

g = acceleration due to gravity = 10m/s²

<em>Substitute these values into equation (i) as follows;</em>

W = 5 x 10 = 50N

<em>Analysis 2: </em>This weight, W, will cause a compression, c, on the spring and this is given according to Hooke's law by;

F = k x c             ---------------(ii)

Where;

F = force acting on the spring = W = 50N

k = the spring's constant = 40N/m

c = compression on the spring

<em>Substitute these values into equation (ii) as follows;</em>

50 = 40 x c

c = \frac{50}{40}

c = 1.25m

Therefore, the compression caused by the load (the masses) is 1.25m

<em>Analysis 3: </em>This compression will cause an elastic energy (E_{E}) in the spring which is given by Hooke's law as follows;

E_{E} = \frac{1}{2} x k x c²      ----------------(iii)

Substitute the values of k and c into equation (iii) as follows;

E_{E} = \frac{1}{2} x 40 x 1.25²

E_{E} = 31.25J

Therefore, the elastic energy produced by the spring is 31.25J

(a) Now, when the upper block (the 3-kg mass) is removed, the 2-kg block moves with some initial velocity (v) thereby resulting into kinetic energy by the 2-kg block. In other words, the elastic energy (E_{E}) in the spring is converted into kinetic energy (K_{E}) by the block. i.e

E_{E} = K_{E}            -------------------------(iv)

Where;

K_{E} = \frac{1}{2} x m x v²            [m = mass of the 2-kg block]

<em>Substitute the value of  </em>K_{E}<em> into equation (iv) as follows;</em>

E_{E} = \frac{1}{2} x m x v²

Where;

E_{E} = 31.25J

m = 2kg

<em>Substitute these values into equation(v) as follows;</em>

31.25 = \frac{1}{2} x 2 x v²

v² = 31.25

v = \sqrt{31.25}

v = 5.59m/s

Therefore, the initial velocity is 5.59m/s. This is also the maximum velocity reached by the block since velocity decreases with elevation such that at maximum height, velocity is zero.

(b) At maximum height, the elastic energy is fully converted to potential energy (P_{E}). i.e

E_{E} = P_{E}            -----------------(vi)

Where;

P_{E} = m x g x h      [m = mass of the block, g = gravity, h = height reached]

Equation (vi) then becomes;

E_{E} = m x g x h             --------------(vii)

<em>Substitute the values of </em>E_{E}<em>, m, and g = 10m/s² into equation (vii) as follows;</em>

31.25 = 2 x 10 x  h

h = \frac{31.25}{20}

h = 1.56m

Therefore, the maximum height reached by the 2-kg block is 1.56m

Kipish [7]3 years ago
6 0

Explanation:

Below is an attachment containing the solution

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import java.awt.event.ActionEvent;

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import java.io.BufferedReader;

import java.io.IOException;

import java.io.InputStreamReader;

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import javax.swing.JPanel;

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// Declaare and define the class MyGUI

abstract class MyGUI extends JFrame implements ActionListener {

JTextField userInput;

JLabel inputDescLbl, resultLbl;

JPanel inputPanel, resultPanel;

JButton evlBtn;

Stack<Object> stk;

// Define the constructor MyGUI

MyGUI() {

super("Tree Address Generator");

inputPanel = new JPanel(new FlowLayout());

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userInput = new JTextField(20);

inputDescLbl = new JLabel("Enter Postfix Expression:");

evlBtn = new JButton("Construct Tree");

evlBtn.addActionListener(this);

resultLbl = new JLabel("Infix Expression:", SwingConstants.LEFT);

add(inputPanel);

add(resultPanel);

inputPanel.add(inputDescLbl);

inputPanel.add(userInput);

inputPanel.add(evlBtn);

resultPanel.add(resultLbl);

stk = new Stack<Object>();

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}

//Stack.java:

// Declare and define the class Stack

class Stack {

private int[] a;

private int top, m;

public Stack(int max) {

m = max;

a = new int[m];

top = -1; }

public void push(int key) {

a[++top] = key; }

public int pop() {

return (a[top--]); }

}

// Declare and define the class Evaluation()

class Evaluation {

public int calculate(String s) {

int n, r = 0;

n = s.length();

Stack a = new Stack(n);

for (int i = 0; i < n; i++) {

char ch = s.charAt(i);

if (ch >= '0' && ch <= '9')

a.push((int) (ch - '0'));

else if (ch == ' ')

continue;

else {

int x = a.pop();

int y = a.pop();

switch (ch) {

case '+':

r = x + y;

break;

case '-':

r = y - x;

break;

case '*':

r = x * y;

break;

case '/':

r = y / x;

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default:

r = 0;

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a.push(r);

}

}

r = a.pop();

return (r);

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}

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// Import packages

import java.util.Scanner;

import java.util.Stack;

// Declare and define the class PostfixToInfix

class PostfixToInfix {

// Determine whether the character entered is an operator or not

private boolean isOperator(char c) {

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public String convert(String postfix) {

Stack<String> s = new Stack<>();

for (int i = 0; i < postfix.length(); i++) {

char c = postfix.charAt(i);

if (isOperator(c)) {

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String a = s.pop();

s.push("(" + a + c + b + ")");

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Scanner sc = new Scanner(System.in);

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System.out.println("Infix : " + obj.convert(postfix));

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}

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e Console X terminated PostfixTolnfix [Java Application] C:\Program Files\Java\jrel.8.0_121\bin\javaw.exe Postfix : ABD++C-D/ .

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B) ratio before operation = 1.55

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A_o/A = 1/(1 - 0.2)

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ε = In(A_o/A)

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ε = 0.2231

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