A block with mass `0.50 kg` is forced against a horizontal spring of negligible mass , compressing the spring a distance of `0.20 m` (figure). When released, the block moves on a horizontal table top for `1.00N//m`. What is the coefficient of kinetic friction `mu _(k)`, between the block and the block and the table?

Options

A. `0.40`
B. `0.50`
C. `0.25`
D. none of these

Answer

A

Explanation

The initial and final kinetic energies are both zero, so the work done by the spring is the negative of the work done by friction or `(1)/(2) kx^(2) = mu _(K) mgl` where `l` is the disance the block moves. Solving for`mu_(K)`.
`mu_(K) = ((1//2) kx^(2))/(mgl) ((1//2) (100 //m) (0.20 m)^(2))/((0.50 kg) (10 m//s^(2)) (1.00m)) = 0.40`