a math question

>>274958

Force?
Force is independant from time, the 3 s are irrelevant.

Force needed to overcome gravitaional pull plus force needed acelerate the mass. Acceleration is needed to be estimated. So

m*(g+a)

>>274967
>the 3 s are irrelevant.
>Acceleration is needed to be estimated
These statements are contradictory.

The most important equation in Newtonian physics is:
Net Force = Mass * Acceleration

If the 185kg weight is sitting on the ground, and lifted up against gravity, the force is 185*9.8 [to counter gravity] + 185*(.8/3) [the movement upward]

I don't actually remember if g or kg is the default SI unit here so your order of magnitude may vary

>>274967
thanks for the response! physics isn't my strong point obv

if i estimate my speed as 0.2666ms^-1 at the end of the lift (and zero at the beginning)

then my average acceleration can be found from (change in speed)/(change in time)
so 0.26/3 = 0.08889ms^-2

am i correct in interpreting the formula you gave as being mass*(gravity*acceleration) ?

so 185 * (9.8*0.0889)
=185*0.871122
=161N

that seems low..

>>274975
kg is the default SI unit

if i follow through the formula you've given
(185*9.8)+(185*(0.8/3))
=1813+(185*0.2666)
=1813+49.321
=1862.321N

that seems like the right sort of figure, can you confirm its right?

>>274976

Yes. The formula for graviational pull is F=m*g and the overall one for a force is F=m*a, I just put it together F=ma+mg = m(a+g)

Your value seems low because you MULTIPLIED a and g, not ADDED them.
185*(9.81+0.098)
=1833 N

Why would it be so naffling that the FORCE needed to lift an object is little more than the one needed to over come its gravitational one?
Or are you thiking of the WORK (Force over time)?
Why would it

>>274976
reread >>274975
>185*9.8 + 185*(0.09)
>185*9.9
>1832 N

>>274984
>>274975
>185*(.8/3)
It's wrong, length divided by time gives you speed, not acceleration.
When you do 0.8/3 you just get your average speed on your lift.

To resume, with actuals informations, you can just say that the force supplied is superior to 185*g at the beginning of the lift.

There are two assumptions/estimations to take into account while solving this question

1)if net acceleration throughout is zero,then the force u apply to lift the weight is the same as the weight of object,I.e,GpF=185*9.8 N
2)if net acceleration is not zero throughout(and force applied is constant,I.e,the ideal case)---->
s=0.8m
a=net acceleration
t=0.3s
u=0(initial velocity of weight)
m=185kg
F=force u applied

s=ut + 1/2*a*t*t
0.8=0 + 1/2*a*9
a=1.6/9

Now,using newtons second law....
F-mg=ma
F=mg+ma
F=185*(9.8+(1.6/9))
F=1845.9N,which is the force u applied while lifting the weight.

Plz do tell me if I'm wrong.....

>>275042
Yeah, if the acceleration is constant through the lift, it's right, but i think it's not likely the case.
However it's a good approximation i think.