A box is sliding up an incline that makes an angle of 17.0° with respect to the horizontal. The coefficient of kinetic friction between the box and the surface of the incline is 0.180. The initial speed of the box at the bottom of the incline is 1.40 m/s. How far does the box travel along the incline before coming to rest?
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To find the distance the box travels along the incline before coming to rest, we can use the principles of kinematics and the equation for the net force acting on the box.
First, let's break down the forces acting on the box along the incline. We have the gravitational force pulling the box downward, which can be separated into two components: the force parallel to the incline (mg.sinθ) and the force perpendicular to the incline (mg.cosθ).
The force of kinetic friction, which opposes the motion of the box, can be calculated by multiplying the coefficient of kinetic friction (μk) by the normal force (mg.cosθ). In this case, the normal force is equal to mg.cosθ because the incline is smooth and there is no vertical acceleration.
Now, using Newton's second law along the incline, we can write the equation for the net force:
Net force = force parallel - force of friction
Since there is no acceleration along the incline (the box comes to rest), the net force is equal to zero:
0 = force parallel - force of friction
Substituting the equations for force parallel and force of friction:
0 = mg.sinθ - μk * mg.cosθ
Now we can solve for the distance the box travels before coming to rest.
The force parallel is equal to the mass of the box (m) multiplied by the acceleration due to gravity (g) multiplied by the sine of the angle (θ):
force parallel = m * g * sinθ
The force of friction is equal to the coefficient of kinetic friction (μk) multiplied by the normal force (mg.cosθ):
force of friction = μk * mg * cosθ
Substituting the equations into the net force equation:
0 = m * g * sinθ - μk * mg * cosθ
Since the mass of the box (m) and the acceleration due to gravity (g) are common, we can cancel them out:
0 = sinθ - μk * cosθ
Now we can solve for the angle (θ):
sinθ = μk * cosθ
Dividing both sides by cosθ:
tanθ = μk
Taking the inverse tangent of both sides:
θ = arctan(μk)
We can plug in the values given:
θ = arctan(0.180)
Using a calculator, we get:
θ ≈ 10.297°
Now that we know the angle of the incline (θ), we can use it to find the distance (d) the box travels along the incline before coming to rest using the formula:
d = (initial velocity)^2 / (2 * acceleration)
The acceleration along the incline is equal to g * sinθ:
acceleration = g * sinθ
Plugging in the values:
acceleration = 9.8 m/s^2 * sin(10.297°)
Using a calculator, we get:
acceleration ≈ 1.705 m/s^2
Now, we can plug in the values of initial velocity (1.40 m/s) and acceleration (1.705 m/s^2) into the distance formula:
d = (1.40 m/s)^2 / (2 * 1.705 m/s^2)
Calculating, we find:
d ≈ 0.693 m
Therefore, the box travels approximately 0.693 meters along the incline before coming to rest.