A particle moves on a rough horizontal ground with some initial velocity say v₀. If $\frac{3}{4}$ of its kinetic energy is lost in friction in time t₀ then coefficient of friction between the particle and the ground is :-
Options

Question

A particle moves on a rough horizontal ground with some initial velocity say v₀. If $\frac{3}{4}$ of its kinetic energy is lost in friction in time t₀ then coefficient of friction between the particle and the ground is :-

Options

G1 Physics · Accepted Answer

Losing 3 4 \frac{3}{4} 4 3 ​ of kinetic energy leaves K(f)=K(i)/4, so v(f)=v₀/2. With constant deceleration µg, v₀-µgt₀=v₀/2, giving µ=v₀/(2gt₀). Hence A.

A particle moves on a rough horizontal ground with some initial velocity say v₀. If 3 4 \frac{3}{4} 4 3 ​ of its kinetic energy is lost in friction in time t₀ then coefficient of friction between the particle and the ground is :-

A particle moves on a rough horizontal ground with some initial velocity say v₀. If 3 4 \frac{3}{4} 4 3 of its kinetic energy is lost in friction in time t₀ then coefficient of friction between the particle and the ground is :-