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)
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)
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).
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.
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.