Question

A box is sliding down an incline tilted at a 11.1° angle above horizontal. The box is initially sliding down the incline at a speed of 1.60 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:s=0.68 m

Step-by-step explanation:

Given

Inclination
`\theta =11.1^(\circ)`

Speed of block(u)=1.6 m/s

Coefficient of kinetic Friction
`\mu _k=0.39`

deceleration provided by friction=g\sin \theta -\mu _kg\cos \theta [/tex]

Using
`v^2-u^2=2as`

Final velocity v=0

`0-1.6^2=2(g\sin \theta -\mu _kg\cos \theta )s`

`s=(-1.6^2)/(2\cdot (9.8\sin 11.1-0.39* 9.8* \cos 11.1))`

s=0.68 m