Question

A particle moves on a rough horizontal ground with some initial velocity say v₀. If 3/4^th of its kinetic energy is lost in friction in time t₀. Then coefficient of friction between the particle and the ground is-

• A. $\frac { \mathrm { v } _ { 0 } } { 2 \mathrm { gt } _ { 0 } }$
• B. $\frac { \mathrm { v } _ { 0 } } { 4 \mathrm { gt } _ { 0 } }$
• C. $\frac { 3 \mathrm { v } _ { 0 } } { 4 \mathrm { gt } _ { 0 } }$
• D. $\frac { \mathrm { v } _ { 0 } } { \mathrm { gt } _ { 0 } }$

Answer 1

Answer: (A)

$\frac { \mathrm { v } _ { 0 } } { 2 \mathrm { gt } _ { 0 } }$

Answer 2

$\frac { 3 } { 4 }$ ^th energy is lost i.e. $\frac { 1 } { 4 }$ KE is left. Hence new velocity becomes $\frac { \mathrm { v } _ { 0 } } { 2 }$ under the retardation µg in time t₀,

$\frac { \mathrm { v } _ { 0 } } { 2 }$ = v₀ – µg t₀ Ž µ = $\frac { \mathrm { v } _ { 0 } } { 2 \mathrm { gt } _ { 0 } }$