Newton's Laws, blocks of mass and string

AI Thread Summary
The discussion centers around a physics problem involving two blocks connected by a string, with one block hanging off a table. The initial calculations presented incorrectly assume constant velocity instead of accounting for acceleration due to gravity. To find the distance between the landing points of the blocks, it's necessary to determine the velocity of each block when the second block leaves the table. Applying Newton's second law to each block individually is essential for solving the problem accurately. The conversation emphasizes the importance of correctly interpreting forces and motion in the context of the problem.
crhscoog
Messages
16
Reaction score
0

Homework Statement



Two small blocks each of mass m, are connected by a string of constant length 4h and negligible mass. Block A is placed on a smooth tabletop and block B hangs over the edge of the table. The tabletop is a distance 2h above the floor. Block B is a distance h above the floor. Block B is then released from rest at a distance h above the floor at time t=0.

The Attempt at a Solution



h= 2h
g= 9.8m/s^2
x= ?

Fnet = ma
mg = 2ma
a= g/2

x= 1/2at^2
h= 1/2(g/2)t^2
t^2 = 4h/g
t= 2sqr(h/g)

v= at
v= (g/2)(2sqr(h/g)

x= vt
= (g/2)(2sqr(h/g)(2sqr(h/g)
= 4hg/2g
= 2h

Is that right? I was wondering if the 2h distance that Block A has to travel before going off the table had a factor although it should be going at constant velocity...
 
Physics news on Phys.org
crhscoog said:

Homework Statement



Two small blocks each of mass m, are connected by a string of constant length 4h and negligible mass. Block A is placed on a smooth tabletop and block B hangs over the edge of the table. The tabletop is a distance 2h above the floor. Block B is a distance h above the floor. Block B is then released from rest at a distance h above the floor at time t=0.

The Attempt at a Solution



h= 2h
g= 9.8m/s^2
x= ?

Fnet = ma
mg = 2ma
a= g/2

x= 1/2at^2
h= 1/2(g/2)t^2
t^2 = 4h/g
t= 2sqr(h/g)

v= at
v= (g/2)(2sqr(h/g)

x= vt
= (g/2)(2sqr(h/g)(2sqr(h/g)
= 4hg/2g
= 2h

Is that right? I was wondering if the 2h distance that Block A has to travel before going off the table had a factor although it should be going at constant velocity...
Since you have not posted the full question, I am not sure what you have been asked to find. However, I can tell you that your solution for whatever you have been asked to find is incorrect since you assumed that the force acting on each block is 1/2m, which is not necessarily the case. In your final section you use the equation x=vt, which assumes that the blocks are traveling with a constant velocity, whereas earlier you assume acceleration (which is correct, the blocks do accelerate).

Start by posting the full question and we can go from there.
 
Ah, I didn't realize that I didn't post the question itself...

"Determine the distance between the landing points of the two blocks"
 
crhscoog said:
Ah, I didn't realize that I didn't post the question itself...

"Determine the distance between the landing points of the two blocks"
Okay, so the first thing you need to do is determine the velocity of each block at the point where the second block leaves the table. This section of the question is very similar to an ideal pulley question so you need to apply Newton's second law to each block individually, resulting in two equations.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Back
Top