Solve Physics Problem: Blocks and Distance

[plane]

i have done some work on this question and was wondering if i have done it right. i would appreciae corrections if I am wrong or atleast mistakes being pointed out. thanks.

two small blocks each of mass 4.0kg are connected by a string of constant length 120cm and negligible mass. Block A is placed on a frictionless tabletop, and block B hangs over the edge of the table. the tabletop is 60cm above the floor. Block B is then released from rest at a height of 30cm above the floor at t=0.

a) find acceleration
b) determine distance between landing points of two blocks

a) Fnet=ma
Fnet=T
T=4a
4g-T=4a
4g-(4a)=4a
4g=8a
a=4(9.8)/8
a=4.9m/s2

b) vf2= vi2+2ad
vf2=0+ 2(4.9)(0.6)
vf=2.4m/s
vfy=3.43m/s
vf=vi+ at
3.43=9.8t
t= 0.35s

dx=vixt+0.5gt2
dx=vixt+0.5gt2
dx=2.4(0.35)+0.5(9.8) (0.35)2
dx= 1.44m

 

[VietDao29]

a) Fnet=ma b) vf2= vi2+2ad dx=vixt+0.5gt2
Fnet=T vf2=0+ 2(4.9)(0.6) dx=2.4(0.35)+0.5(9.8)
T=4a vf=2.4m/s (0.35)2
4g-T=4a dx= 1.44m
4g-(4a)=4a vfy=3.43m/s
4g=8a vf=vi+ at
a=4(9.8)/8 3.43=9.8t
a=4.9m/s2 t= 0.35s

Honestlt, it's very hard to read...
Here's some part that I don't really get what you mean:
F_net = T. What's T? Is it the tension of the string? If you mean the tension of the string then that's incorrect. The force that makes the system accelerate is the weight of the block at the edge of the table.
However, your answer for a seems correct.
-----------------
For b, the problem asks 'determine distance between landing points of two blocks', and you come up with t = 0.35s!
When the block at the edge of the table is still in the air, the whole system is accelerating. But when the block touches the floor. There's no more tension in the string. So the block on the table will travel at the constant velocity (it's the velocity it gets right at the moment the block at the edge of the table touches the floor).
And when that block reach the edge of the table, it falls off the table (with initial velocity). It takes the block 0.35 s to reach the floor (you get this correctly). And while it's still in the air, it continues travel in the Ox direction.
The x-component of its velocity does not change.
Can you calculate how far it lands from the landing point of the first block?
Viet Dao,

 

[plane]

Honestlt, it's very hard to read...
Here's some part that I don't really get what you mean:
F_net = T. What's T? Is it the tension of the string? If you mean the tension of the string then that's incorrect. The force that makes the system accelerate is the weight of the block at the edge of the table.
However, your answer for a seems correct.
-----------------
For b, the problem asks 'determine distance between landing points of two blocks', and you come up with t = 0.35s!
When the block at the edge of the table is still in the air, the whole system is accelerating. But when the block touches the floor. There's no more tension in the string. So the block on the table will travel at the constant velocity (it's the velocity it gets right at the moment the block at the edge of the table touches the floor).
And when that block reach the edge of the table, it falls off the table (with initial velocity). It takes the block 0.35 s to reach the floor (you get this correctly). And while it's still in the air, it continues travel in the Ox direction.
The x-component of its velocity does not change.
Can you calculate how far it lands from the landing point of the first block?
Viet Dao,

i rearranged it, the computer messed them all up. how does it look now?

 

[VietDao29]

I think you should go back and read what I write again.
You got the correct answer for a). However what's T, is it the tension? F_net is not the tension of the string. Here the weight of the mass at the edge of the table makes the whole system accelerate! So what's F_net?
For b/ it's wrong...

b) vf2= vi2+2ad
vf2=0+ 2(4.9)(0.6)

Why d = 0.6 m? After the block at the edge of the table touches the floor, the block on the tabletop travels at the constant velocity. From that moment on, there's no tension to accelerate the block! So what's d?

dx=vixt+0.5gt2
dx=vixt+0.5gt2
dx=2.4(0.35)+0.5(9.8) (0.35)2
dx= 1.44m

After knowing the block takes 0.35 s to touch the floor, you need to calculate how far in the Ox axis (horizontal) the block has traveled. You seem to know it, but it happens to look like you are calculating something else rather than d_x.
Since the x-component of the velocity does not change. So d_x = v_xt.
Viet Dao,