An object , constrained to move along the x-axis is acted upon by a force F(x)=ax + bx^2 where a=5N/m , b=-2m/s
The object is observed directly from x=1m to x=3.0 m... How much work is done by the force?
F(x)= dU/dx, du= F(x)dx
=> dU= (ax + bx²)dx. Integrating we get
U= ax²/2 + bx³/3. At x = 1m we have
U= a/2 + b/3.....(1). At x = 3m we have
U= 9a/2 + 27b/3.....(2). Subtracting we hv
U= 8a/2 + 26b/3. Solving we get
6U= 24a + 52b. Substituting the values of a & b, we have
6U= 120 - 104 => 6U =16, U = 2.67 joules
Final answer
Work=2.67J
2.67j
integral of F(x) dx
= a x^2/2 + b x^3/3
at 3 this is
9a/2 + 27b/3
at 1 it is
a/2 + b/3
subtract
8a/2 - 26b/3
now put in a and b
40/2 + 52/3
etc
Well, well, well, let's calculate the work done by this curious force F(x)=ax + bx^2. To find the work done, we can use the formula:
Work = ∫(F(x)dx),
where we integrate the force over the distance traveled. In this case, the distance traveled is from x=1m to x=3m.
Let's find the antiderivative of the force function:
∫(ax + bx^2) dx = 0.5ax^2 + (b/3)x^3 + C,
where C is a constant of integration.
Now, let's evaluate the definite integral over the given limits:
Work = ∫(from x=1m to x=3m) (0.5ax^2 + (b/3)x^3) dx
= [(0.5a(3)^2 + (b/3)(3)^3) - (0.5a(1)^2 + (b/3)(1)^3)]
= [(0.5a(9) + (b/3)(27)) - (0.5a + (b/3))]
= [4.5a + 9b - 0.5a - (b/3)]
= 4a + 9b - (b/3).
Now, don't get scared by the numbers; let's plug in the values. We're given a = 5 N/m and b = -2 m/s. Let's substitute those values in:
Work = 4(5) + 9(-2) - (-2/3)
= 20 - 18 + 2/3
= 2 + 2/3
= (2/3) + 2/3
= 4/3.
So, the work done by the force is 4/3 units of "I just did some fancy math" measurement.
To find the work done by the force, we need to integrate the force function over the displacement.
The work done by a force F(x) over a displacement Δx is given by the equation:
Work = ∫(F(x) * dx)
In this case, the force function is F(x) = ax + bx^2, where a = 5 N/m and b = -2 m/s. The displacement of the object is from x = 1 m to x = 3.0 m.
Integrating the force function over this displacement, we can find the work done.
Work = ∫((ax + bx^2) * dx)
To calculate the integral, we need to split it into two parts:
Work = ∫(ax dx) + ∫(bx^2 dx)
The first integral is straightforward:
∫(ax dx) = (1/2)a(x^2) + C1
Now, let's calculate the second integral:
∫(bx^2 dx) = (1/3)b(x^3) + C2
where C1 and C2 are constants of integration.
Now, evaluate the definite integrals from x = 1 m to x = 3.0 m:
Work = [(1/2)a(x^2) + C1]│[1, 3.0] + [(1/3)b(x^3) + C2]│[1, 3.0]
Substituting the values:
Work = [(1/2)(5 N/m)(3.0 m^2) + C1 - (1/2)(5 N/m)(1 m^2) - C1] + [(1/3)(-2 m/s)(3.0 m^3) + C2 - (1/3)(-2 m/s)(1 m^3) - C2]
Simplifying further, we get:
Work = [(15/2) N + C1 - (5/2) N] + [-6 m^2/s + C2 + 2 m^2/s]
C1 and C2 are the constants of integration, which cancel out when taking the difference:
Work = (15/2 - 5/2) N + (-6 + 2) m^2/s
Work = 5 N + (-4) m^2/s
Therefore, the work done by the force is 5 N + (-4) m^2/s.
Note: The work is measured in joules (J), which is the unit of energy.