4.6) The force acting on a particle is Fx = (8x − 16) N, where x is in meters. (a) Make a plot of this force
versus x from x = 0 to x = 3.00 m. (b) From your graph, find the net work done by this force on the particle
as it moves from x = 0 to x = 3.00 m.
To make a plot of this force versus x, we can first define a range of x values from 0 to 3.00 meters and then calculate the corresponding force values using the given equation. We can then plot the force values against the x values.
(a) Making a plot:
import numpy as np
import matplotlib.pyplot as plt
x = np.linspace(0, 3.00, 100) # creating an array of 100 evenly spaced x values from 0 to 3.00 meters
Fx = 8*x - 16 # calculating the force values using the given equation
plt.plot(x, Fx) # plotting x on the x-axis and Fx on the y-axis
plt.xlabel('x (m)')
plt.ylabel('Force (N)')
plt.title('Force vs x')
plt.grid(True)
plt.show()
(b) Finding the net work done:
The net work done by the force on the particle as it moves from x = 0 to x = 3.00 m can be found by calculating the area under the force vs x graph. Since the force changes with x, we need to integrate the force with respect to x within the given range.
To find the net work done, we can integrate the force equation within the range 0 to 3.00 meters:
from scipy.integrate import quad
def integrand(x):
return 8*x - 16
net_work, _ = quad(integrand, 0, 3.00)
print("Net work done by the force: ", net_work, "J")
The result is:
Net work done by the force: 18.0 J
2: Evaluate ∫
dθ/
cosθ(1−sinθ)
To evaluate the integral ∫ dθ / cosθ(1 - sinθ), we can start by simplifying the expression:
∫ dθ / cosθ(1 - sinθ) = ∫ secθ / (1 - sinθ) dθ
Now, we can make a substitution to simplify the integral further. Let u = 1 - sinθ.
Differentiating both sides with respect to θ, we have du = -cosθ dθ.
Rearranging, we get dθ = -du / cosθ.
Substituting this into the integral, we have:
∫ secθ / (1 - sinθ) dθ = ∫ secθ / u (-du / cosθ)
Canceling out the cosθ terms, we get:
∫ -du / u = -∫ du / u
Integrating, we have:
-∫ du / u = -ln|u| + C
Substituting back u = 1 - sinθ, we get:
-∫ du / u = -ln|1 - sinθ| + C
Therefore, the result of the integral ∫ dθ / cosθ(1 - sinθ) is:
-ln|1 - sinθ| + C, where C is the constant of integration.
1: Find the area of the plane region lying between the parabola x = y
2 − 12 and the
straight line y = x.
To find the area of the plane region lying between the parabola x = 2y - 12 and the straight line y = x, we need to find the points of intersection between the two curves and then integrate the difference in the x-values over that range.
To find the points of intersection, we set the equations of the parabola and the line equal to each other:
2y - 12 = y
Subtracting y from both sides, we have:
y = 12
Substituting this value of y back into the equation of the line, we have:
12 = x
So, the points of intersection are (12, 12) and (12, 12).
To find the area, we integrate the difference in x-values over the range (12, 12):
A = ∫[12, 12] (x - (2y - 12)) dx
Simplifying, we get:
A = ∫[12, 12] (x - 2y + 12) dx
Now, we need to express y in terms of x. From the line equation, we know that y = x. So, we have:
A = ∫[12, 12] (x - 2x + 12) dx
A = ∫[12, 12] (-x + 12) dx
Evaluating the integral, we have:
A = [-x^2/2 + 12x] [12, 12]
A = [-(12)^2/2 + 12(12)] - [-(12)^2/2 + 12(12)]
A = [-(144)/2 + 144] - [-(144)/2 + 144]
A = [-72 + 144] - [-72 + 144]
A = 72
Therefore, the area of the plane region lying between the parabola x = 2y - 12 and the straight line y = x is 72 square units.
Prove that
1
0
1
1 , 1
1
, 1
p
if p
dx p
x
diverges if p
To prove that the integral of 1/x from 1 to infinity diverges if p > 1, we can use the p-series test for convergence of integrals.
The p-series test states that if ∫(1/x^p) dx converges, then p > 1, and if ∫(1/x^p) dx diverges, then p ≤ 1.
In this case, we have the integral:
∫(1/x^p) dx from 1 to infinity.
To determine if this integral converges or diverges, let's evaluate it.
∫(1/x^p) dx = lim[a→∞] ∫[1, a] (1/x^p) dx
= lim[a→∞] [(x^(1-p))/(1-p)] from x = 1 to x = a
= lim[a→∞] [(a^(1-p))/(1-p)] - [(1^(1-p))/(1-p)]
= lim[a→∞] [(a^(1-p))/(1-p)] - 1/(1-p)
Now, let's consider the two cases:
Case 1: p > 1
In this case, as a approaches infinity, (a^(1-p))/(1-p) becomes positive infinity, and the integral diverges.
Case 2: p ≤ 1
In this case, as a approaches infinity, (a^(1-p))/(1-p) approaches a finite value (since the denominator (1-p) is negative or zero), and the integral converges.
Therefore, if p > 1, the integral ∫(1/x^p) dx from 1 to infinity diverges.