Correct Answer - a. `24m`; `6ms^-2`, b. `1/8`
(a) At the extreme position blocks stops.
Applying work-energy theorem, we get
`mg sin 37^@=1/2ks^2`
`2xx10xxsxx3/5=1/2xx100xxs^2`
On solving `s=0.24m`
Acceleration at its lowest point, `a=(ks-mg sin 37^@)/(m)`
`=(100xx0.24-2xx10xx3/5)/(2)=6ms^-2`
(b) `W_g+W_(f riction)+W_(spri ng)=DeltaKE`
`mg sin 37^@+mumgcos37^@xxs=1/2ks^2`
`mgsin37^@-1/2ks=mumgcos 37^@`
`2xx10xx3/5-1/2xx100xxs=muxx2xx10xx4/5`
Gives `s=0.020m`
`mu=(12-50s)/(16)impliesmu=1/8`