Question

A block of mass m has initial velocity u having direction +x axis. The block stops after covering distance S causing similar extension in the spring of spring constant K holding it. If μis the kinetic friction between the block and the surface on which it was moving, the distances S is given by

• A. $\frac { 1 } { \mathrm {~K} }$ μ²m²g²
• B. (mKu² -μ²m²g²)^1/2
• C. (μ²m²g² + mKu² + μmg)^1/2
• D. (μ²m²g² – mKu² + μmg)^1/2

Answer 1

Answer: (D)

(μ²m²g² – mKu² + μmg)^1/2

Answer 2

$\frac { 1 } { 2 }$ mu² = - μmgS + $\frac { 1 } { 2 }$ KS²

i.e., S² + $\frac { 2 \mu \mathrm { mgS } } { \mathrm { K } } - \frac { \mathrm { mu } ^ { 2 } } { \mathrm {~K} } = 0$

Solving for μ,

S = (μ²m²g² – mKu² - μmg)