When a particle located a distance x feet from the origin, a force of x^2+2x pounds acts on it. How much work is done in moving it from x=1 to x=3?
W=[1,3]∫(x^2+2x)dx
W=[(x^3)/3+(2x^2)/2]
W=27/3+9-(1/3-1)
W=9+9-1/3-1=50/3pound/feet
Well, it seems like this particle is stuck in a bit of a quadratic force field! To calculate the work done, we can use the formula:
Work = ∫F(x) dx
So, let's integrate the force function F(x) = x^2 + 2x with respect to x, and evaluate it from x=1 to x=3:
Work = ∫(x^2 + 2x) dx
= (1/3)x^3 + x^2 + C
Now, let's plug in the values:
Work = ((1/3)(3)^3 + (3)^2) - ((1/3)(1)^3 + (1)^2)
= (27/3 + 9) - (1/3 + 1)
= 9 + 9 - (1/3 + 1)
= 18 - (4/3)
If we simplify further:
Work = 54/3 - 4/3
= 50/3
So, it appears that the work done in moving the particle from x=1 to x=3 is approximately 16.67 pounds-feet. Now, that's quite a workout for a particle!
To calculate the work done in moving the particle from x=1 to x=3, we need to integrate the force function with respect to x over the given range.
The work done by a force is given by the formula:
Work = ∫F(x) dx
Given that the force acting on the particle at x feet from the origin is F(x) = x^2 + 2x pounds, we can integrate this function over the range from x=1 to x=3.
Work = ∫(x^2 + 2x) dx
To evaluate this integral, we can use the power rule of integration:
∫x^n dx = (1/(n+1)) * x^(n+1) + C
Applying this rule to each term of the integrand:
Work = ∫(x^2) dx + ∫(2x) dx
= (1/3) * x^3 + x^2 + C
To find the work done in moving the particle from x=1 to x=3, we substitute these values into the formula:
Work = [(1/3) * 3^3 + 3^2] - [(1/3) * 1^3 + 1^2]
= [(1/3) * 27 + 9] - [(1/3) * 1 + 1]
= [9 + 9] - [1/3 + 1]
= 18 - 4/3
= 54/3 - 4/3
= 50/3
Therefore, the work done in moving the particle from x=1 to x=3 is 50/3 pounds-feet.
To determine the work done in moving the particle from x = 1 to x = 3, we can use the principles of calculus.
The work done (W) can be obtained by integrating the force function (F) over the displacement (dx) from x = 1 to x = 3. The relationship for work is given by:
W = ∫F dx
Let's start by finding the force function. We are given that the force acting on the particle is x^2 + 2x pounds. Therefore, F = x^2 + 2x.
Next, we need to integrate the force function with respect to x:
∫(x^2 + 2x) dx
To integrate the above expression, we apply the power rule of integration. The integral of x^n is (1/n+1)x^(n+1). Applying this rule, we get:
∫(x^2 + 2x) dx = (1/3)x^3 + (2/2)x^2 + C
where C is the constant of integration.
To evaluate the definite integral from x = 1 to x = 3, we subtract the value of the integral at x = 1 from the value of the integral at x = 3:
W = [(1/3)(3^3) + (2/2)(3^2)] - [(1/3)(1^3) + (2/2)(1^2)]
Simplifying the expression:
W = [(1/3)(27) + (2/2)(9)] - [(1/3)(1) + (2/2)(1)]
W = [9 + 9] - [1/3 + 1]
W = 18 - 4/3
W = 54/3 - 4/3
W = 50/3
Therefore, the work done in moving the particle from x = 1 to x = 3 is 50/3 pounds-feet.