A block of mass `0.18 kg ` is attached to a spring of force constant `2 N//m` The coefficient of friction between the block and the force is `0.1` insitially its block is at rest and the block as spring is an streched , As impalse is given to the block as shown in the figure . The block sides a distance of ` 0.06` in and comes to the first time . The initial velocity of the for blocks is mis `V = N 10` then `N` is .

Options

A. 3
B. 4
C. 5
D. 2

Answer

B

Explanation

Here, `m=0.8 kg, k=2 N//m, mu=0.1`,
`x=0.06m, `
According to work energy theorem , we have
`(1)/(2) m upsilon^(2)=mu mg x+(1)/(2) k x^(2)`
`:. (1)/(2) xx0.18=0.1xx0.18xx10xx0.06`
`+(1)/(2)xx2xx(0.06)^(2)`
or `upsilon^(2)=(144xx10^(-4))/( 9xx10^(-2))=16xx10^(-2)`or`upsilon=(4)/(10) m//s`
As per question, `(N)/(10)=(4)/(10)` or`N=4`