Answer:
The box will cover a distance of 0.9199m before coming to rest
Explanation:
We are given;
Angle of tilt; θ = 12°
Speed of sliding down; u = 1.5 m/s
Coefficient of kinetic friction; μ = 0.34
We are told that the box is sliding down an incline tilted at a 12° angle above horizontal.
Thus,
The components of the weight of the block would be;
Fx = mg sinθ = mg sin 12
Fy = mg cosθ = mg cos 12
For, the normal force on the block, it will be counter balanced by the Y component of weight of block and so we have;
Normal force; Fn = mg cos 12
Now formula for the frictional force would be given by;
Ff = μmg cos 12
So, Ff = 0.34mg cos 12
So, the net force along the inclined plane is;
Fnet = Fx - Ff
Fnet = mg sin 12 - 0.34mg cos 12
Where Fnet = mass x acceleration.
Thus;
ma = mg sin 12 - 0.34mg cos 12
m will cancel out to give;
a = g sin 12 - 0.34g cos 12
a = 9.81(0.2079) - 0.34(9.81 × 0.9781)
a = -1.223 m/s²
According to Newton's equation of motion, we have;
(v² - u²) = 2as
s = (v² - u²)/2a
Final velocity is zero. Thus;
s = (0² - 1.5²)/(2 × -1.223)
s = -2.25/-2.446
s = 0.9199 m
Thus, the box will cover 0.9199m before coming to rest
