# How much work would it take to push a # 4 kg # weight up a # 12 m # plane that is at an incline of # pi / 6 #?

I found:

I considered the following diagram:

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To calculate the work done, you can use the formula:

[ W = F \cdot d \cdot \cos(\theta) ]

Where:

- ( W ) is the work done,
- ( F ) is the force applied (in the direction of motion),
- ( d ) is the displacement,
- ( \theta ) is the angle between the force and the displacement.

Given:

- ( m = 4 , \text{kg} ) (mass),
- ( d = 12 , \text{m} ) (distance),
- ( \theta = \pi/6 ) (angle).

To find ( F ), we need to decompose the weight force into its components parallel and perpendicular to the incline.

[ F = mg \sin(\theta) ]

[ F = 4 , \text{kg} \times 9.8 , \text{m/s}^2 \times \sin(\pi/6) ]

[ F = 4 \times 9.8 \times 0.5 ]

[ F = 19.6 , \text{N} ]

Now, we can calculate the work done:

[ W = 19.6 , \text{N} \times 12 , \text{m} \times \cos(\pi/6) ]

[ W = 19.6 \times 12 \times \frac{\sqrt{3}}{2} ]

[ W = 19.6 \times 12 \times \frac{\sqrt{3}}{2} ]

[ W = 19.6 \times 6\sqrt{3} ]

[ W \approx 202.812 , \text{J} ]

So, it would take approximately ( 202.812 , \text{J} ) of work to push the 4 kg weight up a 12 m plane that is at an incline of ( \pi/6 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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