CATAGORY: Mechanics
QUESTION: A block is put on a horizontal table with which it has a
coefiicient of friction of 0.25.  While on theis table the block is pushed
up against a spring such that the spring is compressed and the block is
stationary.  The block is now released and it slides 60 cm across the
table before coming to a stop.  Now the same experiment is repeated except
that the spring is compressed twice as much as before.  What is the
distance the block slides across the table in this new case?

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Some interpretation first I assume you mean 60cm AFTER the block
clears the spring.  When the spring is compressed a distance X the
block slides 60cm

|
|            before                   after
|  _   _   _   _                        _
|_/ \_/ \_/ \-|_|--X--|-----60cm-------|_|
---------------|------|-----------------|
                      0

The energy put into the block by the spring is .5 k X^2 Where k is the
spring constant.  The energy drain on the block from friction is Uk
Mg(X+.6m) where Uk is the coeficient of sliding friction, M is the
mass of the block anfd g is the acceleration due to gravity.  From
conservation of energy we have

(1)    Uk Mg (X + .6m) = .5 k X^2   

                      
For the new arrangement we have a similar equation except X -> 2X
and .6m -> L

|
|            before                       after
|       _                                   _
|/\/\/\|_|-----2X-----|----------L=?-------|_|
---------------|------|---------------------|
                      0
(2)   Uk mg (2X + L) = .5 k (2X)^2

We have Uk =.25 but we dont have M, k, X, or L which we are trying to
find.  We have only two equations and three unknowns.  :-( Unless we
make further assumptions we can not go further.  Suppose that the
energy lost from friction can be neglected while the pring is still
pushing.  Presumably the distance is so short compared to the length
of the free sliding part that we can ignore it.  If so then we have

(3)   Uk M g .6m = .5 k X^2

(4)      ---> 1.2m Uk M g/k =X^2
and

(5)   Uk M g L = .5 k (2X)^2 

(6)      ---> L = .5 k (2X)^2 / Uk M g

Substitution of (4) into (6) gives

(7)   L = .5 k 4(1.2 Uk M g/k) / Uk M g

        = .5 x 4 x 1.2m
        
        =  2.4m (or four times as far)
        
         ************
         *          *
         * L = 2.4m *           
         *          * Approximate answer
         ************