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:
[tex]\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\\[/tex]
Where:
[tex]\mu[/tex] 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
[tex]\theta = 30^0\\\mu = 0.25\\g = 9.8m/s^2[/tex]
Required
acceleration of the box
Substitute the given parameters into the resulting expression above:
Recall that:
[tex]gsin\theta - \mu g cos\theta = a_x\\[/tex]
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
