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
The question is understood as below:
Block of mass
1 Answer
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
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
PE of the spring
Force of friction
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=(-2mumg+-sqrt((2mumg)^2 -4xxK(-mv^2)))/(2K)#
#S=1/K(-mumg+-sqrt((mumg)^2 +Kmv^2))#
Ignoring the