Question

A box is placed on a 30o frictionless incline. What is the acceleration of the box as it slides down the incline

Answer 1

Answer:

2.78m/s²

Explanation:

Complete question:

A box is placed on a 30° frictionless incline. What is the acceleration of the box as it slides down the incline when the co-efficient of friction is 0.25?

According to Newton's second law of motion:

$\sum F_x = ma_x\\F_m - F_f = ma_x\\mgsin\theta - \mu mg cos\theta = ma_x\\gsin\theta - \mu g cos\theta = a_x$

Where:

$\mu$ is the coefficient of friction

g is the acceleration due to gravity

Fm is the moving force acting on the body

Ff is the frictional force

m is the mass of the box

a is the acceleration'

Given

$\theta = 30^0\\\mu = 0.25\\g = 9.8m/s^2$

Required

acceleration of the box

Substitute the given parameters into the resulting expression above:

Recall that:

$gsin\theta - \mu g cos\theta = a_x$

9.8sin30 - 0.25(9.8)cos30 = ax

9.8(0.5) - 0.25(9.8)(0.866) = ax

4.9 - 2.1217 = ax

ax = 2.78m/s²

Hence the acceleration of the box as it slides down the incline is 2.78m/s²