Question

a box of mass m slides across a horizontal surface with initial kinetic energy k. the coefficient of kinetic friction between the box and the surface is μ. which of the following expressions correctly describes the time the box will slide across the surface before coming to rest?

Explanation:

Step1: Relate K to initial velocity

The initial kinetic energy is $K = \frac{1}{2}mv_0^2$, so solve for $v_0$:
$$v_0 = \sqrt{\frac{2K}{m}}$$

Step2: Find friction-induced acceleration

Kinetic friction force $f_k = \mu mg$, so acceleration $a = -\frac{f_k}{m} = -\mu g$ (negative for deceleration).

Step3: Use kinematic equation for time

Final velocity $v=0$, use $v = v_0 + at$. Substitute values:
$$0 = \sqrt{\frac{2K}{m}} - \mu g t$$
Solve for $t$:
$$t = \frac{1}{\mu g}\sqrt{\frac{2K}{m}} = \sqrt{\frac{2K}{\mu^2 m g^2}} = \sqrt{\frac{2K}{m\mu^2 g^2}}$$

Answer:

$$\boldsymbol{t = \frac{1}{\mu g}\sqrt{\frac{2K}{m}}}$$
(or equivalently $\boldsymbol{\sqrt{\frac{2K}{m\mu^2 g^2}}}$)