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1. This is a higher physics question and i have gotten part a correct but i have done something wrong in part b? The answer to part a is 226N
2. (Original post by #engineer)
This is a higher physics question and i have gotten part a correct but i have done something wrong in part b? The answer to part a is 226N
Can you post what you've got so far?
3. (Original post by Muttley79)
Can you post what you've got so far?
Sorry my pen smugged
4. (Original post by #engineer)

Sorry my pen smugged
Try resolving the new force of 140N then you know the friction from part (a).

Then use F = ma remembering what 'F' in the formula represents.
5. (Original post by Muttley79)
Try resolving the new force of 140N then you know the friction from part (a).

Then use F = ma remembering what 'F' in the formula represents.
Which formula should I use because I used Fx=Fcos(theta)-Ff... And i got it wrong?
6. (Original post by #engineer)
This is a higher physics question and i have gotten part a correct but i have done something wrong in part b? The answer to part a is 226N
Resolve the new forces horizontally and subtract the friction (calculated in a)

Then use F=ma
7. (Original post by ValerieKR)
Resolve the new forces horizontally and subtract the friction (calculated in a)

Then use F=ma
That gets an acceleration of 1.37 but ghe answer is 0.371
8. (Original post by #engineer)
That gets an acceleration of 1.37 but ghe answer is 0.371
Then you've made a mistake somewhere
9. (Original post by #engineer)
That gets an acceleration of 1.37 but ghe answer is 0.371
(2*140cos(20degrees)-226)/100 = 0.3711393382
10. Attachment 578638578642
(Original post by ValerieKR)
Then you've made a mistake somewhere
Attached Images

11. (Original post by ValerieKR)
(2*140cos(20degrees)-226)/100 = 0.3711393382
I am such an idiot sorry I was writing 126 instead of 226
12. (Original post by #engineer)
Attachment 578638578642

You're subtracting 126 instead of 226

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