Question

A box is sliding down an incline tilted at a 11.6° angle above horizontal. The box is initially sliding down the incline at a speed of 1.10 m/s. The coefficient of kinetic friction between the box and the incline is 0.390. How far does the box slide down the incline before coming to rest?

Answer 1

Answer:

0.34 m

Explanation:

We are given that

$\theta=11.6^{\circ}$

Initial speed of box=u=1.1 m/s

Coefficient of friction,$\mu=0.39$

We have to find the distance slide down by the box before coming to rest.

Friction force=f=$\mu mg=$

Where $g=9.8 m/s^2$

Net force=$mgsin\theta-\mu mg$

$ma=mg(sin\theta-\mu cos\theta)$

$a=g(sin\theta-\mu cos\theta)$

Substitute the values

$a=9.8(sin11.6-0.39cos11.6)=9.8(-0.18)=-1.76 m/s^2$

$v=0$

$v^2-u^2=2as$

Substitute the values

$0-(1.1)^2=2(-1.76)s$

$-3.52s=-(1.1)^2$

$s=\frac{-(1.1)^2}{-3.52}$

$s=0.34 m$

Hence, the box slide down the incline 0.34 m before coming to rest.