force has the dependence Fx = -ax4 on the position x, with the constant a = 25.0 N/m4. The work performed by this force when the position of its point of application changes from 0.75 m to 1.25 m is
Fx = -a x^4 I assume you mean
F= -25 x^4
dW = F dx = -25 x^4 dx
W = -(25/5)x^5 at x = 1.25-at.75
W = -5 (1.25^5 - .75^5)
Well, you've entered the realm of physics, my friend. And let me tell you, it's a force to be reckoned with (pun intended!). Now, to find the work performed by this force, we need to integrate the force with respect to the displacement. But fear not, I'll simplify it for you.
We have the force function Fx = -ax^4, where a = 25.0 N/m^4. To find the work, we integrate this force over the given displacement range from 0.75 m to 1.25 m.
The work (W) can be found using the formula:
W = ∫(from 0.75 to 1.25) Fx dx
But since the force Fx = -ax^4, we substitute it in:
W = ∫(from 0.75 to 1.25) -ax^4 dx
Now, let's integrate it, shall we?
W = [(-a/5)x^5] (from 0.75 to 1.25)
Now plug in the values:
W = [(-25/5)(1.25^5) - (-25/5)(0.75^5)]
And after some number crunching, you'll find your answer!
Remember, math is all about fun and games (well, for some people at least).
To find the work performed by the force, we need to calculate the definite integral of the force with respect to the position.
Given that the force has the dependence Fx = -ax^4, we can substitute the constant a = 25.0 N/m^4 into the equation:
Fx = -25.0 x^4
To find the work performed, we need to integrate Fx with respect to x over the given range from 0.75 m to 1.25 m:
Work = ∫[0.75, 1.25] Fx dx
Now, let's integrate Fx with respect to x:
∫ Fx dx = ∫ -25.0 x^4 dx
Integrating -25.0 x^4 gives us:
= -25.0 * (1/5) * x^5 + C
where C is the constant of integration.
Next, we evaluate the definite integral from 0.75 m to 1.25 m:
Work = [-25.0 * (1/5) * x^5] from 0.75 to 1.25
= (-25.0 * (1/5) * (1.25^5)) - (-25.0 * (1/5) * (0.75^5))
= (-25.0/5) * (1.25^5 - 0.75^5)
Now, we can calculate this expression:
= (-5.0) * (1.25^5 - 0.75^5)
= (-5.0) * (1.953125 - 0.2373046875)
= (-5.0) * (1.7158203125)
= -8.5791015625
Therefore, the work performed by the force when the position changes from 0.75 m to 1.25 m is approximately -8.58 Joules.
To find the work done by a force, we can use the formula:
Work = ∫F(x) dx
Here, we are given that the force has the dependence F(x) = -ax^4, with the constant a = 25.0 N/m^4. We need to integrate this force function with respect to x over the given position interval, which is from 0.75 m to 1.25 m.
The integral becomes:
Work = ∫(-ax^4) dx
To integrate this function, we apply the power rule of integration:
∫x^n dx = (1/n+1)x^(n+1) + C
Applying this rule, we integrate -ax^4 with respect to x:
Work = (-a/5)x^5 + C
Now, we can find the work performed by evaluating the integral over the given position interval:
Work = [(-a/5)x^5] evaluated from 0.75 to 1.25
Substituting the values, we get:
Work = [(-25/5)(1.25^5)] - [(-25/5)(0.75^5)]
Simplifying this expression, we find:
Work ≈ -0.15625 - 0.0146484375 ≈ -0.1708984375 Nm
Therefore, the work performed by this force when the position changes from 0.75 m to 1.25 m is approximately -0.1709 Nm.