`1 fm = __micrometer`

Answer

A, B, C, D

Explanation

Given `k=100BN/m, m=1kg and F=10 N
a. In the equilibrium position
`Compression `=delta=F/k=10/100`
`=0.1m=10cm`
b. The below imparts a speed of 2 m/s to the block towards left.
`:.+P.E.+K.E.=1/2kdelta^2+1/2Mv^2`
`=1/2x100xx(0.1)^2+1/2x1xx4`
`=0.5+2=2.5J`
`c. Time period `2pisqrt(M/k)`
`=2pi sqrt(M/100)=pi/5 sec`
d. Let the amplitude be x whigh means the distasnce betwen the mean position and the extreme position.
So in the extreme position, compression of the spring is `(x-K)`
Since in SHM the totl energy remains constant,
`1/2k(x+delta)^2=1/2kdelta^2+1/2mv^2+Fx`
`=2.5+10x
[because 1/2kdelta^2+1/2mv^2=2.5]`
so, `50(x+0.1)^2=2.5+10x`
`:. 50x^2+0.5+10x=2.5+10x`
`=:. 50x^2=2`
`rarr x^2=2/50=4/100`
`rarr x=2/50x=20m`
e. potential energy at the left extreme is given by
`P.E. =1/2k(x+delta)^2`
`=1/2x100xx(0.1+0.2)^2`
`=50xx(0.09)=4.5J`
f. Potential energy at the right extreme is given by
`P.E. 1/2k(x+delta)^2-F(2x)`
`[2x`=distance between two extreme)`
`=4.5-10(0.4)=0.5J`
The different values is b, e, and f do not violate law of conservastion of energy as the work is done by the exterN/Al force 10 N.