# Terminal Speed question

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#1
Trying to figure this out I dont know where Im going wrong

A steel ball of mass 0.15kg released from rest in a liquid, falls a distance of 0.20m in 5.0s. Assuming the ball reaches terminal speed within a fraction of a second, calculate

i) Its terminal speed

ii) the drag force on it when it falls at terminal speed

I got up to now

S = 0.20
U = 0
V =
A =
T = 5

Using S = (u+v)/2 * t

I got 0.20 = (0 + v)/2 *5
0.40 = (0+v)*5
v = 0.08 ms^-1

This is wrong as the answer is 0.04ms^-1

0
6 years ago
#2
(Original post by AhmedDavid)
Trying to figure this out I dont know where Im going wrong

A steel ball of mass 0.15kg released from rest in a liquid, falls a distance of 0.20m in 5.0s. Assuming the ball reaches terminal speed within a fraction of a second, calculate

i) Its terminal speed

ii) the drag force on it when it falls at terminal speed

I got up to now

S = 0.20
U = 0
V =
A =
T = 5

Using S = (u+v)/2 * t

I got 0.20 = (0 + v)/2 *5
0.40 = (0+v)*5
v = 0.08 ms^-1

This is wrong as the answer is 0.04ms^-1

Terminal speed is a constant speed. No acceleration.
You don't use suvat equations. Just velocity = distance / time

If the velocity is constant there is no resultant force.
What is the downwards force equal to in this case?
The drag force is the upwards force.
0
#3
(Original post by Stonebridge)
Terminal speed is a constant speed. No acceleration.
You don't use suvat equations. Just velocity = distance / time

If the velocity is constant there is no resultant force.
What is the downwards force equal to in this case?
The drag force is the upwards force.
Yeah thanks I know how to work out the answer using velocity = distance / time

& I also know how to do part b)

However why doesnt it work with suvats
0
6 years ago
#4
(Original post by AhmedDavid)
Yeah thanks I know how to work out the answer using velocity = distance / time

& I also know how to do part b)

However why doesnt it work with suvats
Suvats are equations for uniform acceleration. This was no acceleration.

If you want to look at it closer, then a=0 in the suvats gives you uniform velocity and they simplify to velocity = distance / time

v=u+at with a = 0 gives final velocity = initial velocity. That is, same constant speed.

s = t(u+v)/2 which you used was used incorrectly because you put initial velocity as 0 and final as v when there was no change in velocity.

Try it with u=v and you get

s = t(v+v)/2
which gives
s=vt
which is the standard equation for constant speed.
So it will work with suvat but there's no point.
You used the suvat incorrectly.
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