Question

A block of mass M slides along a horizontal table with speed $v_0.$ At x=0 it hit's 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.

• A. $\frac{bg{V}_0^2{M}^2}{(k+Mg)}$
• B. $\frac{bg{V}_0^2{M}^2}{(k+bMg)}$
• C. $\frac{bg{V}_0^2{M}^2}{2(k+bMg)}$
• D. $\frac{V_0^2{M}^2}{2(k+bMg)}$

Answer 1

Answer: (C)

$\frac{bg{V}_0^2{M}^2}{2(k+bMg)}$

The retarding forces acting on the block at some coordinate $x$ are $kx$ and $\mu Mgx$

When the block comes at rest at some distance $x_0$ from $x=0$, all the kinetic energy goes into the heat produced by friction until that point and potential energy of spring attained.

$⟹\frac{1}{2}M{v}_0^2=\frac{1}{2}k{x}_0^2+{\int }_0^{x_0}bMgxdx$

$=\frac{1}{2}k{x}_0^2+\frac{1}{2}bMg{x}_0^2$

$⟹x_0=\sqrt{\frac{M{v}_0^2}{k+2bMg}}$
Loss in mechanical energy is due to friction only,

Thus loss in mechanical energy=$\frac{1}{2}bMg{x}_0^2$

$=\frac{bg{v}_0^2{M}^2}{2(k+bMg)}$