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# I am a little confused with this physics question watch

1. I'm a little confused about this question I'll show what I've got so far please point out if I'm wrong but I did

I) 0.5*600*(9.5/cos(30))^2 = 36100J
ii) 600*9.81*4.1 = 24132.6J
iii) this is where I'm confused I understand work done is force*distance so the distance I worked out to be 4.1/sin(30) =8.2m
Then I need to work out the force due to friction but idk how is it 600*9.81*sin (30)or am I completely wrong
Also the last question I do think know where to start

Any help would be appreciated thanks😀
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2. for i it says kinetic energy travelling towards the slope, you have calculated kinetic energy when on the slope
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3. (Original post by Hasham123)
for i it says kinetic energy travelling towards the slope, you have calculated kinetic energy when on the slope
Ahaha okay yeah thanks
4. https://youtu.be/qwIUqtITHEE
This video explains the other bit
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5. (Original post by Hasham123;7602737423)
https://youtu.be/qwIUqtITHEE
This video explains the other bit
So would it be 2943*8.2*cos(180)
6. yes
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7. (Original post by Hasham123)
yes
Thanks a million 😀😀
8. (Original post by Amberinho)
I'm a little confused about this question I'll show what I've got so far please point out if I'm wrong but I did

I) 0.5*600*(9.5/cos(30))^2 = 36100J
(Original post by Hasham123)
for i it says kinetic energy travelling towards the slope, you have calculated kinetic energy when on the slope
Actually you wouldn't do that calculation for KE on the slope because KE isn't a vector quantity.
1kg moving at 1m/s has the same KE as 1kg moving at -1m/s (or moving at 1m/s in any other direction)

note 0.5*600*9.5^2 = 27075 J the KE the trolley had on the level
where did the extra 9025 J of KE suddenly come from when it started going up the slope?

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