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A block of mass $m$ sliding on a smooth horizontal surface with a velocity $v$ meets horizontal spring of spring constant $k$ fixed with wall. Find maximum compression of spring.

Why is the word maximum used to describe the compression? How does it differ from, for example-minimum compression?

The compression of a spring itself IMPLIES that it be maximum, because if not then NO compression can occur because this would mean the v of the block is 0.

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  • $\begingroup$ Maximum compression is when speed of block will be zero.The spring will generate a force to stop the block(or to stop compression) that will decelerate the block $\endgroup$ – user227513 Apr 13 at 18:12
  • $\begingroup$ Just a guess, but I think it means that you should assume that there no losses of KE from the block to friction, so all the energy of the block goes into spring compression and nothing else. $\endgroup$ – Ubaid Hassan Apr 13 at 18:15
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    $\begingroup$ Because asking "what is the compression of the spring" is unanswerable because the spring's compression changes over time $\endgroup$ – Aaron Stevens Apr 13 at 21:11
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$ "MAXIMUM" $ by it the author to the problem means that what is the largest compression in the spring before all the kinetic energy of the block is converted into elastic potential energy . The obvious answer to minimum compression is $ 0 $ ( the state in which the spring is in its mean position ) .You are wrong if it wasn't maximum compression then velocity of the mass wouldn't be $ 0 $ you can see it is quite obvious that the total energy of the system is shared between the E.P.E and the K.E initially it is purely kinetic in nature but there must be a moment when it is purely elastic and at that moment as. Spring 's potential energy is $ \frac {1}{2} kx² $ hence $ x $ must be maximum .

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The maximum possible compression of the spring happens when all of the mass's kinetic energy has been stored in the spring as elastic potential energy- that is why this point in the mass's travel is special. This point occurs when the spring has brought the mass to a complete stop.

At that point, the energy stored in the spring is $\frac{1}{2}kx^2$ where k is the spring constant and x is the compression. This is equal to the kinetic energy of the mass before it hit the spring, which is $\frac{1}{2}mv^2$ where m is the mass and v is its velocity.

Take care with units when you set up the equation.

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