Solving Spring & Block Homework Problem

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The discussion revolves around solving a physics problem involving two blocks connected by a spring on a smooth horizontal surface. The key points include the need to apply energy conservation principles rather than relying solely on tension forces, as the acceleration of block A is not constant. The correct method involves calculating the spring's extension and equating the energy stored in the spring with the kinetic energy of the blocks and the potential energy of block B. The final velocity of block A when it leaves the surface is determined to be 1.5 m/s, highlighting the importance of including the kinetic energy of both blocks in the calculations. Clarification of variables and the complete problem statement is emphasized for effective assistance.
Suraj M
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Homework Statement


WIN_20150226_194753.JPG

The figure has 2 blocks, each 320 g connected by a light string. The horizontal surface is smooth. The block A is attached to a spring of spring constant 40 N/m whose other end is fixed 40cm above the horizontal. Initially the spring is vertical and unstretched.

Homework Equations


E= ½kx²
+basic

The Attempt at a Solution


I considered the point where the block leaves the surface.
WIN_20150226_195849.JPG

so $$kx \sin(\theta) =mg$$
taking ##\sin(\theta) = \frac{0.4}{0.4+ x} ##
i get x = 0.1 cm so from that i get s = 0.3 cm
also acceleration of block A is given by
$$a=\frac{ M_B g}{M_A+M_B}$$
so i get ## v = sqrt{2as}##
v = √3
what I'm concerned is the spring has a component of force along the horizontal, but its variable, but also it effects the velocity, the given answer is 1.5 m/s.how do i consider the horizontal component?
 
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Wrote the whole text of the problem please and explain you notations. What is x? What is the question?
 
Suraj M said:
also acceleration of block A is given by
$$a=\frac{ M_B g}{M_A+M_B}$$

The acceleration of block A is not constant . The result you have used is obtained if tension were the only force acting on it . There is something else you haven't taken into account .

Instead use energy conservation .
 
ehild said:
Wrote the whole text of the problem ?
I don't know what you mean, I typed the whole question as given.
And yes I am sorry x is the extension produced in the spring as shown in the diagram.
 
Tanya Sharma said:
The acceleration of block A is not constant . The result you have used is obtained if tension were the only force acting on it . There is something else you haven't taken into account .

Instead use energy conservation .
Ok I understand, but I still need x right, is the method I found x, right?
using that x, I should find the energy stored in the spring then add to the kinetic energy of the block and equate to change in potential energy(blockB), is this right? Or is there anything else I should be including? And thank you
 
Well, from post #1 we know the answer is 1.5 m/s, from post #4 we know what x is, but we still don't know the question !

Suraj M said:
I don't know what you mean, I typed the whole question as given.
And yes I am sorry x is the extension produced in the spring as shown in the diagram.
 
Suraj M said:
Ok I understand, but I still need x right, is the method I found x, right?

I am not sure ,what method you have used to find x . You can use Pythagoras Theorem to find x.

Suraj M said:
using that x, I should find the energy stored in the spring then add to the kinetic energy of the block and equate to change in potential energy(blockB), is this right?

Right .

I may have understood the question , but you need to write the complete question so that other members can understand your work and provide help.
 
BvU said:
but we still don't know the question !
Oh I'm sorry i didn't realize. Anyway the last part of the question is to find the velocity of the object A at the instant it leaves the horizontal plane.
Tanya Sharma said:
I am not sure ,what method you have used to find x
By this way
Suraj M said:
I considered the point where the block leaves the surface. https://physicsforums-bernhardtmediall.netdna-ssl.com/data/attachments/62/62693-dc7a39c502a2c1b87185def92b37bdda.jpg
so
kxsin(θ)=mg​
kx \sin(\theta) =mg
taking sin(θ)=0.40.4+x\sin(\theta) = \frac{0.4}{0.4+ x}
i get x = 0.1 m so from that i get s = 0.3 cm
Ok tanya, i tried the method you suggested:
$$½kx² +½mv²=m_Bgs$$
where s is the distance moved by Block A along the horizontal.
from the above equation i get 2.17 m/s but the answer is 1.5 m/s mathematical error? i doubt it.
 
Last edited by a moderator:
0.1 cm isn't very much. You certain ?
 
  • #10
it should be 0.1 m . I have used that in the further calculations. not 0.1cm.
 
  • #11
Suraj M said:
$$½kx² +½mv²=m_Bgs$$

The second term is incorrect .

Suraj M said:
from the above equation i get 2.17 m/s but the answer is 1.5 m/s mathematical error? i doubt it.

1.5 m/s is the correct answer .
 
  • #12
Oh ok got it , i didn't consider the KE of block B. Thanks a lot all 3 you.
 
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