Question

A certain scientist collected Force data at the following accelerations, a, for each mass shown, m a: 1, 5, 10, 15, 20, 25, 30, 35,40, 45 m: 1, 5, 10, 15 The units of acceleration are and those of mass are kg (the units of Force are N). 1. Knowing that Fma, determine the Force vectors associated with each mass studied 2. Plot Force versus acceleration for each mass studied. All plots must be on the same graph with the following appearance characteristics: One curve is blue with diamonds One curve is green with squares One curve is yellow with stars One curve is red with triangles A legend located in the upper left hand corner of the graph Properly labeled axes with appropriate units A descriptive title

Answer 1

MATLAB Code:

close all
clear
clc

% Exercise 5
fprintf('Exercise 5\n-------------------------------------------------------------\n')
a = [1 5 10 15 20 25 30 35 40 45];
m = [1 5 10 15];

F = zeros(length(m), length(a)); % Each row corresponds to a mass
disp('Force Vectors:')
colors = ['b', 'g', 'y', 'r'];
styles = ['d', 's', '*', '^'];
figure, hold on
for i = 1:length(m)
F(i,:) = m(i)*a;
fprintf('\tFor mass m = %d:\n\t', m(i)), disp(F(i,:))
plot(a, F(i,:), sprintf('-%s%s',colors(i), styles(i)))
end
xlabel('a (m/s^2)'), ylabel('F (N)'), title('F = ma for different masses')
legend('m = 1', 'm = 5', 'm = 10', 'm = 15', 'Location', 'northwest')
hold off

fprintf('\nExercise 6\n-------------------------------------------------------------\n')
M = 0.001:0.001:0.01; % Mass
E = [1.0 1.6 2.9 3.6 4.4 5.7 6.4 7.3 8.1 10]*1e14; % Energy

fprintf('INFO: polyfit(...) fits the data points to a polynomial of desired degree using least squares approach.\n\n')

% E = m*c^2 => E = a*m and we need to find a (polynomial of degree 1)
p = polyfit(M,E,1);
a = p(1);
fprintf('Obtained Model: E = (%e)*m\n', a)
fprintf('Speed of light, c: %e (m/s)\n', sqrt(a))

MM = linspace(min(M), max(M), 100);
EE = a*MM;
figure, plot(M,E/1e14,'o',MM,EE/1e14), xlabel('Mass (kg)'), ylabel('Energy (J) x 1e14')
legend('Data Points', 'Polynomial Fit', 'Location', 'northwest')
title('E = mc^2')

Output:

Exercise 5
-------------------------------------------------------------
Force Vectors:
   For mass m = 1:
   1 5 10 15 20 25 30 35 40 45
   For mass m = 5:
   5 25 50 75 100 125 150 175 200 225
   For mass m = 10:
   10 50 100 150 200 250 300 350 400 450
   For mass m = 15:
   15 75 150 225 300 375 450 525 600 675

Exercise 6
-------------------------------------------------------------
INFO: polyfit(...) fits the data points to a polynomial of desired degree using least squares approach.

Obtained Model: E = (9.587879e+16)*m
Speed of light, c: 3.096430e+08 (m/s)

Plots:

F ma for different masses 700 600 m= 10 m= 15 500 400 300 200 100 0 5 10 15 202530 35 4045 a (m/s)

E=mc2 10 O Data Points Polynomial Fit 7 0 4 8 9 10 × 10-3 3 4 7 Mass (kg)