We have some object hanging at the end of a vertical pole. Someone moves the pole so the object's height off the Earth is h(t)=100+(16t-2t^3)cos(2*π*t). The object has mass 4.3kg. Find the force on the object from the pole at time equals 5 seconds.

To find the force on the object from the pole at time \( t = 5 \) seconds, we need to calculate the object's acceleration using the second derivative of its height function \( h(t) \), and then apply Newton's second law of motion \( F = ma \), where \( m = 4.3 \) kg is the mass of the object, and \( a \) is the acceleration at time \( t = 5 \) seconds.

The given height function is:
\[
h(t) = 100 + (16t - 2t^3) \cos(2 \pi t)
\]
### Step 1: Differentiate the height function to find the velocity and acceleration.

#### First derivative (velocity):
Using the product rule, the derivative of \( h(t) \) with respect to \( t \) is:

\[
v(t) = \frac{d}{dt} h(t) = \frac{d}{dt} \left[ (16t - 2t^3) \cos(2 \pi t) \right]
\]

We will use the product rule: 
\[
\frac{d}{dt} \left[ u(t) \cdot v(t) \right] = u'(t) \cdot v(t) + u(t) \cdot v'(t)
\]
where \( u(t) = (16t - 2t^3) \) and \( v(t) = \cos(2 \pi t) \).

Now let's compute \( v(t) \) and \( a(t) \) for \( t = 5 \).The acceleration of the object at \( t = 5 \) seconds is approximately \( -60 + 680 \pi^2 \) m/s².

Now, using Newton's second law \( F = ma \), with the mass \( m = 4.3 \) kg:

\[
F = 4.3 \times (-60 + 680 \pi^2)
\]

The force exerted on the object by the pole at \( t = 5 \) seconds is approximately:

\[
F = -258 + 2924 \pi^2 \, \text{N}.
\]

If you want further calculations or simplifications to get a numerical value, feel free to ask!

Would you like to proceed with that or need more details? Here are some related questions:

1. How can we use this method to find the object's velocity at \( t = 5 \) seconds?
2. How do product and chain rules combine when differentiating composite functions?
3. Can we determine the maximum or minimum forces exerted on the object?
4. What would happen to the force if the object's mass was doubled?
5. How does the acceleration at \( t = 5 \) compare to other points in time?

**Tip**: Always check if units are consistent in force calculations to avoid errors.

This solution calculates the force on an object hanging from a moving pole using the height function h(t)=100+(16t-2t^3)cos(2πt). Learn how to apply differentiation and Newton's second law to find the force at t=5 seconds for an object with a mass of 4.3kg. Suitable for undergraduate physics or calculus students.

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