How Is Kinetic Friction Affecting the Pulley System with Different Masses?

[braindead101]

The block with mass m2=0.50kg is found to have a speed of 0.30m/s after it has dropped 0.80m. how large a (kinetic) friction force retards the motion of the block with mass m1=2.0kg?

(side note from teacher)
To solve this problem, you can use a string constraint. the total length of the string (L) is constant. However, as block m1 moves to the right, the portion of the string that is horizontal shortens by delta y and this length is distributed over the portion of the string taht is vertical according to the relationship; |delta x|=1/2|delta y|. By taking a derivative with respect to time twice, we have a1=2a2.

i think i did this completely wrong

i drew a free body diagram for each,for 1st free body diagram (mass1) and found m1 had Fn of 19.6

on the 2nd free body diagram (mass 2), i had Fg-Ft = ma
then, 4.9 N - 2Ft =0 , Ft = 2.45, then i said the Ft in the first free body diagram was also equal to 2.45, and therefore the kinetic friction force equals 2.45

its probably wrong, since i didnt use any info my teacher gave in the side note

 

[Fermat]

...
i drew a free body diagram for each,for 1st free body diagram (mass1) and found m1 had Fn of 19.6
on the 2nd free body diagram (mass 2), i had Fg-Ft = ma
then, 4.9 N - 2Ft =0 ,...

You put the accln of m2 as zero - it should be a2.

 

[braindead101]

ohh.
okay, i did it like u said, but now I am stuck

for the mass 1 fbd, i got
Ft - Ff = a1

for the mass 2 fbd, i got
Fg - 2Ft = ma2 (is it 2Ft?)
then isolate for a2
4.9 - 2Ft = 0.5a2
9.8 - 4Ft = a2

and the equaion from fbd #1:
Ft - Ff = ma1
since a1 = 2a2
Ft - Ff = 4a2 (1)
then i substituted a2 into equation (1)
Ft - Ff = 4(9.8 - 4Ft)
Ft - Ff = 39.2 - 16Ft
17Ft = 39.2 + Ff

okay I am stuck, how do i find Ft?

 

[Fermat]

ohh.
okay, i did it like u said, but now I am stuck
for the mass 1 fbd, i got
Ft - Ff = a1
for the mass 2 fbd, i got
Fg - 2Ft = ma2 (is it 2Ft?)
then isolate for a2
4.9 - 2Ft = 0.5a2
9.8 - 4Ft = a2
and the equaion from fbd #1:
Ft - Ff = ma1
since a1 = 2a2
Ft - Ff = 4a2 (1)
then i substituted a2 into equation (1)
Ft - Ff = 4(9.8 - 4Ft)
Ft - Ff = 39.2 - 16Ft
17Ft = 39.2 + Ff
okay I am stuck, how do i find Ft?

You have,

9.8 - 4Ft = a2 (from fbd #2)
Ft - Ff = 4a2 (from fbd #1)

eliminate Ft from these two eqns, giving you,

9.8 - 4Ff = 17a2

Now use the info about the movement of m2 given in the question to work out a2.
Substitute for a2 and solve for Ff.

 

[braindead101]

so, a2 = 0.30m/s / 0.80m = 0.375 m/s^2
and then sub it into get Ff = 0.856 N
oops, that's not the acceleraton
hmm, u need kinematics??
is the veloctiy given v1 or v2
a = 9.8 m/s^2
d= 0.80
if velocity given is v2, so v1 = 0?

 

[braindead101]

actually i have a question, i used 2Ft in one of the equations, and I am wondering if this is right.

for this step i used 2Ft :
for the mass 2 fbd, i got
Fg - 2Ft = ma2 (is it 2Ft?)

 

[Fermat]

actually i have a question, i used 2Ft in one of the equations, and I am wondering if this is right.
...

Yes, that's right.

so, a2 = 0.30m/s / 0.80m = 0.375 m/s^2 ...

Nope, that's wrong
use the kinematic eqn,
v² = u² + 2as