Spring and block

1. Feb 26, 2015

Suraj M

1. The problem statement, all variables and given/known data

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.

2. Relevant equations
E= ½kx²
+basic

3. The attempt at a solution
I considered the point where the block leaves the surface.
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?

2. Feb 26, 2015

ehild

Wrote the whole text of the problem please and explain you notations. What is x? What is the question?

3. Feb 26, 2015

Tanya Sharma

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 .

4. Feb 26, 2015

Suraj M

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.

5. Feb 26, 2015

Suraj M

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

6. Feb 26, 2015

BvU

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 !

7. Feb 26, 2015

Tanya Sharma

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

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.

8. Feb 27, 2015

Suraj M

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.
By this way
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: May 7, 2017
9. Feb 27, 2015

BvU

0.1 cm isn't very much. You certain ?

10. Feb 27, 2015

Suraj M

it should be 0.1 m . I have used that in the further calculations. not 0.1cm.

11. Feb 27, 2015

Tanya Sharma

The second term is incorrect .

1.5 m/s is the correct answer .

12. Feb 27, 2015

Suraj M

Oh ok got it , i didn't consider the KE of block B. Thanks a lot all 3 you.

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