A box is sliding down an incline tilted at an angle 14.0° above horizontal. The box is sliding down the incline at a speed of 1.70 m/s. The coefficient of kinetic friction between the box and the incline is 0.380. The box slide at 1.16 m down the incline before coming to rest.
The diagrammatic expression of the question given can be seen in the image below.
We need to resolve the component of the plane into parallel and perpendicular forces.
In the parallel direction;
In the perpendicular direction, the normal force N = mg cos θ
The frictional force F = μN
where;
∴
The frictional force F = μ × mg cos θ
Thus, the resultant force in the parallel direction of the plane is:
- F = mg sinθ - μ mg cos θ
- F = sin θ - μ cos θ
Similarly, using the second equation of motion
Force of an object = mass × acceleration
∴
- Total force = mg cos θ - μmg cos θ
- ma = mg cos θ - μmg cos θ
Making acceleration (a) the subject of the formula, we have:
acceleration (a) = g(sin θ - μ cos θ)
- a = 9.8 × (sin 14 - 0.380 ×cos 14)
- a = 9.8 × (0.2419 - 0.380 × 0.9703)
- a = 9.8 × (0.2419 - 0.368714)
- a = - 1.2427 m/s²
Finally, using the kinematics equation;
- [tex]\mathbf{v^2 - u^2 =2as}[/tex]
where;
- v = final velocity
- u = initial velocity
- s = distance of the box
- a = acceleration
[tex]\mathbf{v^2=u^2+2as }[/tex]
[tex]\mathbf{(0)^2=(1.70)^2+2\times (-1.2427)s }[/tex]
[tex]\mathbf{-(1.70)^2=+2\times (-1.2427)s }[/tex]
Making (s) the subject of the formula:
[tex]\mathbf{s = \dfrac{-(1.70)^2}{2 \times -1.2427}}[/tex]
[tex]\mathbf{s = \dfrac{(1.70)^2}{2 \times 1.2427}}[/tex]
[tex]\mathbf{s = \dfrac{2.89}{2.4854}}[/tex]
s = 1.16 m
Therefore, we can conclude that the distance of how far the box slides down the incline before coming to rest is 1.16 m
Learn more about kinetic friction here:
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