Suvat equations help please
I'm very confused. Why do I get 2 different accelerations depending on whether I use v=u+at or v2=u2+2as here ?
Help please
Original post by JanaALEVEL
I'm very confused. Why do I get 2 different accelerations depending on whether I use v=u+at or v2=u2+2as here ?
Help please
Help please
I get 3/4 m/s² both ways. Post your working and we'll try to help you find the mistake.
Original post by MarkFromWales
I get 3/4 m/s² both ways. Post your working and we'll try to help you find the mistake.
Using v=u+at
3=a(4) a=3/4
but using v2=u2+2as
9=2a(8)
a=9/16
Original post by JanaALEVEL
Using v=u+at
3=a(4) a=3/4
but using v2=u2+2as
9=2a(8)
a=9/16
3=a(4) a=3/4
but using v2=u2+2as
9=2a(8)
a=9/16
Apologies. Yes, you are right. I'm going to think about this!
The suvat formulas only apply when acceleration is constant (or zero). I'm thinking about whether this is the case.
I'm thinking that as the box accelerates the frictional force changes so the acceleration isn't constant so suvat formulas don't apply.
Original post by JanaALEVEL
I'm very confused. Why do I get 2 different accelerations depending on whether I use v=u+at or v2=u2+2as here ?
Help please
I'm very confused. Why do I get 2 different accelerations depending on whether I use v=u+at or v2=u2+2as here ?
Help please
Well if you are trying to answer the question of work done, all you need to use is the formula work done= Force * Distance
Original post by Chakram
Well if you are trying to answer the question of work done, all you need to use is the formula work done= Force * Distance
work done by the friction
Original post by JanaALEVEL
work done by the friction
Yes, the friction acts in the opposite direction to the motion so the work done by friction will be negative.
I still don't know why they use the a=9/16
Original post by JanaALEVEL
I'm very confused. Why do I get 2 different accelerations depending on whether I use v=u+at or v2=u2+2as here ?
Help please
I'm very confused. Why do I get 2 different accelerations depending on whether I use v=u+at or v2=u2+2as here ?
Help please
Friction's limiting, varies from 0 to μR; net force/accel over the distance varies, so don't use SUVAT. Besides, why focus on accel if you're looking for work done?
Work done by F is 120*8 = 960 J, yet mass obtains KE = 0.5*20*3^2 = 90 J, so work done by friction is?
(edited 4 years ago)
Original post by JanaALEVEL
I still don't know why they use the a = 9/16
v^2 = u^2 + 2as => (m/2)(v^2 - u^2) = mas = Fs
=> KEfinal - KEstart = Net W => KEbox = WF + WFr
So this SUVAT eqn implies energy eqn and vice versa, so the 'a' that fits will work; average/effective accel over the distance (with corresponding net force).
(edited 4 years ago)
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