Block of mass m has ini. velo. u having drxn +x axis.the blk stops after covering dist. S causing similar ext. in the spring of spring const. K holding. if mu is the k.friction btwn the blck and surface on which it was moving,the distance S is given by?

The question is understood as below:
Block of mass m has initial velocity u having direction along +x axis. The block stops after covering distance S causing similar extension in the spring of spring constant K holding the block. if μ is the coefficient of kinetic friction between the block and surface on which it was moving, the distance S is given by the equation

1 Answer
A08
Jul 27, 2017

I got a different answer than the published result.
Did I make a mistake!

Explanation:

Assuming that initially the spring is in its equilibrium position.

Initial energy of block is its kinetic energy =1/2mv^2

Using Law of Conservation of energy:
When the block stops it initial kinetic energy is converted in to mechanical potential energy of spring which gets stretched by distance S and remaining energy is spent doing work against force of friction during its movement.

PE of the spring=1/2KS^2

Force of friction =mumg

Work done against force of friction ="Force"xx"distance"
=mumgxxS

Equating the initial and final energies we have

1/2mv^2=1/2KS^2+mumgS
=>KS^2+2mumgS-mv^2=0

Solving the quadratic in S we get

S=(-2mumg+-sqrt((2mumg)^2 -4xxK(-mv^2)))/(2K)
S=1/K(-mumg+-sqrt((mumg)^2 +Kmv^2))

Ignoring the -ve root as the movement is in +x direction we get
S=1/K(sqrt((mumg)^2 +Kmv^2)-mumg)

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