A block of mass M slides along a horizontal table with speed `v_(0)`. At `x=0`, it hits a spring with spring constant k and begins to experience a friction force. The coefficient of friction is variable and is given by `mu=bx`, where b is a positive constant. Find the loss in mechanical energy when the block has first come momentarily to rest.
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Options

A. `(v_(0)^(2)M^(2)g)/(2(k-bMg))`.
B. `(v_(0)^(2)M^(2)g)/(2(k+bMg))`.
C. `(2(v_(0)^(2)M^(2)g))/((k+bMg))`.
D. `(2(v_(0)^(2)M^(2)g))/((k+bMg))`.

Answer

B

Explanation

Net retarding force `=kx +bMgx`
`:.` Net retardatopm `=((k+bMg)/M).x`
So, we can write
`v.(dv)/(dx) = -((k+bMtg)/(M)).x`
`v.(dv)/(dx)=-((k+bMg)/M)int_(0)^(x)xdx`
or `int_(v0)^(0)v.dv =-((k+bMg)/M)int_(0)^(x)xdx`
or `x=sqrt(M/(k_bMg))v0`
Loss in vechanical energy
`DeltaE=1/2Mv_(0)^(2)-1/2kx^(2)`
or `DeltaE=1/2 Mv_(0)^(2) -1/2 (m/(k+bMg))v_(0)^(2)`
or `DeltaE=v_(0)^(2)/2[M-k(M/(k+bMg))]`
`=(v_(0)^(2)M^(2)g)/(2(k+bMg))`.