A 0.5 kg block slides from a point A [fig.1.17] on a horizontal track with an initial speed of 3 m.s^-1 towards a weightless horizontal spring of length 1 m and of force constant 2 N.m^-1. The part AB of the track is frictionless and the part BC has coefficients of static and kinetic friction as 0.22 and 0.2 respectively. If the distance AB and BD are 2 m and 2.14 m respectively. Find the total distance covered by the black before it comes to rest (g=10 m.s^-2)

A 0.5 kg block slides from a point A [fig.1.17] on a horizontal track

1.	A 0.5 kg block slides from a point A [fig.1.17] on a horizontal track with an initial speed of 3 m.s^-1 towards a weightless horizontal spring of length 1 m and of force constant 2 N.m^-1. The part AB of the track is frictionless and the part BC has coefficients of static and kinetic friction as 0.22 and 0.2 respectively. If the distance AB and BD are 2 m and 2.14 m respectively. Find the total distance covered by the black before it comes to rest (g=10 m.s^-2)
Answer» Solution :Initial kinetic energy of the block `=1/2mv^2=1/2times0.5times(3)^2=2.25J` The part AB of the horizontal TRACK is frictionless and HENCE during the PASSAGE of the block over this path no energy is LOSS in energy in the part BD =workdone against function in that path `=mumg times BD=0.2times0.5times10times2.14=2.14J` So, the kinetic energy of the blok when it reaches the point D=2.25-2.14=0.11 J Let us ASSUME that the block compresses the spring through an amount x. According to the principle of conservation of energy. work done against friction+work done in compressing the spring =0.11 or`mu mgx+1/2kx^2=0.11` or,`0.2times0.5times10timesx+1/2times2x^2=0.11` or,`x^2+x-0.11=0` or`(x+1.1)(x-0.1)=0` `therefore` x=0.1 m [as x cannot be -1.1m] `therefore` The distance covered by the block before it comes to rest =2+2.14+0.1=4.24m.