A 337kg crate needs to be lifted to a height of 2.3m using a ramp that is 7.6m long. Ideally, how much work will it take to lift the crate?

Answer 1

I found: #~~7596J#

I think that you can use the work done against gravity as: #W_g=mgh# with your data: #W_g=337*9.8*2.3~~7596J#

You can also try using the force needed to push the crate (equal to the #y# component of the weight) and the distance and you'll get the same result.

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Answer 2

Elena Albarran

The work done is ( \text{Work} = \text{Force} \times \text{Distance} ), where ( \text{Force} = \text{Weight of the crate} \times \text{Vertical displacement} ).

( \text{Weight} = \text{mass} \times \text{gravitational acceleration} )

( \text{Work} = (\text{mass} \times \text{gravitational acceleration}) \times \text{Vertical displacement} \times \sin(\theta) ), where ( \theta ) is the angle of the ramp.

The ideal work required to lift the crate is the work done against gravity, taking into account the ramp's angle.

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Answer from HIX Tutor

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.