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Trouble With Springs-2

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Question-
A block mass m = 1kg is placed over another block of mass M = 3kg. M is placed over a smooth surface. Both the blocks move with velocity v towards a spring of constant K = 400 N/m and
μ = 0.2 for the two blocks. Then, the maximum value of v, so that there may not be slipping between m and M is -

My Attempt-
Let me call block m and block M together as the system. When the system compresses the spring and momentarily comes to rest, the compression can be found out through conservation of energy.
         Elastic Potential Energy Gained = Kinetic Energy Lost (Work Done by Friction is Internal)
                                                               1/2 Kx2 = 1/2 (M+m) v2
Plugging in the values,
                                                                       x = v/10
During this compression, Kinetic Friction Force acts on block m and reduces its velocity to zero.
Deceleration due to Kinetic Friction Force is given by,
                                                                       2N = 1kg x a
                                                                             a = 2m/s2
Since this is the maximum frictional force that is acting, the maximum velocity that it can reduce to zero over the same distance x can be obtained from kinematics.
On substituting x = v/10 in the kinematics equation, we get
                                                                       v = 0.4 m/s

I am told that this answer is wrong. Have I committed any mistake?

asked Mar 21, 2017 in Physics Problems by DoubtExpert (340 points)
is the answer 0.2 m/s
Yes! You are Right!

1 Answer

1 vote

Here a two block system moves. Imagine it was colliding with a spring attached and causes a compression. And now it is compressing and a time comes where it has maximum compression , at this time the opposite force to the blocks would be maximum. --------------(1)

Now here the force is variable and hence the acceleration. So what can be the maximum acceleration that friction can provide ? first we need to sort that out.

$\mu = 0.2 $
So $a_{max}= 2 m/s^2$

This $a_{max}$ is critical acceleration for which the two blocks move toghether.It is
that acceleration that maximumally friction can provide so as to oppose the relative motion.
Force required to produce this acceleration $F= 4× 2 =8 N $

Update :
Now since a force corresponding to the $a_{max}$ is 8 N , If the spring in the maximum compressed state would exert maximum force and both blocks survive the force together and the top block does not slide and we have counted here the maximum force all other time that is other than maximum compressed state the force would be less and hence friction would be able to provide that.

This maximum force is provided at time as in ---(1)
Hence applying conservation of energy ,

$F^2/ 2k = 1/2m v^2$
$8×8/ (2× 400) = 1/2 × 4 × v^2$

$v = 0.2 m /s$

answered Mar 21, 2017 by physicsapproval (2,290 points)
edited Mar 21, 2017 by physicsapproval
Which is the force that you mention is variable?
The force due to  the spring
'This a max is critical acceleration for which the two blocks move toghether.' What does it mean?
well;  question asks that the block on top does not slip  $a _{max} $ is that acceleration that maximumally friction can provide so as to oppose the relative motion
Could you tell what mistake I have made?
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