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Edexcel M2 Textbook Question Help

Hey!

m2.png

I can't get the answer to part (b). It's supposed to be 61.2J. The SolutionBank shows a solution to a different question. Please help! :biggrin:

It might be simple but I have a lot of trouble in this chapter. :/

Thank you :smile:
Original post by MineXO
Hey!

m2.png

I can't get the answer to part (b). It's supposed to be 61.2J. The SolutionBank shows a solution to a different question. Please help! :biggrin:

It might be simple but I have a lot of trouble in this chapter. :/

Thank you :smile:


The loss in potential energy will be mg sin 25 x s, where s is the distance the parcel travels down the plane in those 2 seconds. I agree with 61.2J

You should get the acceleration as 1.4771... ms^-2
(edited 10 years ago)
Reply 2
Original post by ghostwalker
The loss in potential energy will be mg sin 25 x s, where s is the distance the parcel travels down the plane in those 2 seconds. I agree with 61.2J

You should get the acceleration as 1.4771... ms^-2


Sorry for not being specific!

I got the acceleration, and the loss I know will be mgsin25 x s, I just can't form the equation :/ I get very lost in this chapter. I thought we might need to use the work-energy principle.
Original post by MineXO
Sorry for not being specific!

I got the acceleration, and the loss I know will be mgsin25 x s, I just can't form the equation :/ I get very lost in this chapter. I thought we might need to use the work-energy principle.


You don't need to use the work-energy principle, but you can, and that would give you.

Loss in PE = work done against friction + gain in kinetic energy.

You'd still need to work out the distance travelled (suvat).
Reply 4
Original post by ghostwalker
You don't need to use the work-energy principle, but you can, and that would give you.

Loss in PE = work done against friction + gain in kinetic energy.

You'd still need to work out the distance travelled (suvat).


Thank youu!!! :biggrin: :biggrin: & I really need to work on this chapter :/

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