Newton's Laws, blocks of mass and string

In summary: Fnet = mamg = 2maa= g/2x= vth= 1/2(g/2)t^2t^2 = 4h/gt= 2sqr(h/g)v= atv= (g/2)(2sqr(h/g)x= vt = (g/2)(2sqr(h/g)(2sqr(h/g) = 4hg/2g = 2hSo, the distance between the two blocks is 2h.
  • #1
crhscoog
17
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...
 
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  • #2
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.
 
  • #3
Ah, I didn't realize that I didn't post the question itself...

"Determine the distance between the landing points of the two blocks"
 
  • #4
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.
 

1. What are Newton's Laws of Motion?

Newton's Laws of Motion are three fundamental principles of physics that describe the motion of objects. The first law states that an object at rest will remain at rest, and an object in motion will remain in motion at a constant velocity unless acted upon by an external force. The second law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The third law states that for every action, there is an equal and opposite reaction.

2. How do Newton's Laws apply to blocks of mass connected by a string?

Newton's Laws apply to blocks of mass connected by a string in the same way they apply to any other objects. The first law still holds true, as the blocks will remain at rest or in motion unless acted upon by an external force. The second law applies to each individual block, where the net force acting on the block is equal to its mass times its acceleration. The third law also applies, as the forces exerted by each block on the other are equal and opposite.

3. What is the relationship between mass and acceleration in Newton's second law?

According to Newton's second law, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This means that for a given force, a smaller mass will result in a larger acceleration, and a larger mass will result in a smaller acceleration.

4. Can an object have a net force of zero and still be in motion?

Yes, an object can have a net force of zero and still be in motion if it was initially set in motion by an external force and is moving at a constant velocity. This is because the first law of motion states that an object in motion will remain in motion at a constant velocity unless acted upon by an external force.

5. How does friction affect the motion of objects connected by a string?

Friction can have a significant impact on the motion of objects connected by a string. It can act as an external force and change the velocity or acceleration of the connected objects. Additionally, friction can cause the string to stretch, which can affect the tension and forces between the objects. Taking friction into account is important when applying Newton's Laws in scenarios involving blocks of mass connected by a string.

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