A box is sliding up an incline that makes an angle of 21.8° with respect to the horizontal. The coefficient of kinetic friction between the box and the surface of the incline is 0.170. The initial speed of the box at the bottom of the incline is 3.48 m/s. How far does the box travel along the incline before coming to rest?
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To find out how far the box travels along the incline before coming to rest, we need to analyze the forces acting on the box.
Let's break down the forces into components parallel and perpendicular to the incline:
The force of gravity can be divided into two components:
- The component perpendicular to the incline (mg * cosθ), where m is the mass of the box and g is the acceleration due to gravity.
- The component parallel to the incline (mg * sinθ).
The normal force (N) acts perpendicular to the incline.
The force of kinetic friction (fk) opposes the motion and acts parallel to the incline. The magnitude of the force can be calculated as fk = μk * N, where μk is the coefficient of kinetic friction.
Since the box is sliding up the incline, the force of kinetic friction acts in the downward direction.
Now, let's consider the net force acting on the box parallel to the incline:
Net force = force parallel to incline - force of kinetic friction
At the start, the net force is equal to the force parallel to the incline:
Net force (at start) = mg * sinθ
As the box moves up the incline, the force of kinetic friction opposes its motion, gradually reducing its speed until it comes to a stop.
When the box comes to rest, the net force will be zero:
Net force (at rest) = 0
By equating the two net force equations, we can find the distance traveled by the box before coming to rest:
mg * sinθ = μk * N
To find the normal force (N), we need to consider the vertical forces:
Sum of vertical forces = Net vertical force = 0
mg * cosθ - N = 0
Now, we have two equations involving N. Let's solve them simultaneously:
mg * cosθ - N = 0 (Equation 1)
mg * sinθ = μk * N (Equation 2)
From Equation 1, we can express N in terms of mg:
N = mg * cosθ
Now substitute N in Equation 2:
mg * sinθ = μk * (mg * cosθ)
Simplifying, we get:
sinθ = μk * cosθ
Now, divide both sides by cosθ:
tanθ = μk
Finally, solve for θ:
θ = atan(μk)
Plug in the given coefficient of kinetic friction (μk = 0.170) to find θ:
θ = atan(0.170)
Using a calculator, we find the angle θ is approximately 9.786°.
Now, we can use the angle and the initial speed of the box to calculate the distance traveled using the kinematic equation:
vf^2 = vi^2 + 2as
Where:
- vf is the final velocity (0 m/s since the box comes to rest)
- vi is the initial velocity (3.48 m/s)
- a is the acceleration (which can be calculated using the component of gravity parallel to the incline)
The acceleration can be calculated as:
a = g * sinθ
Plug in the values:
a = 9.8 m/s^2 * sin(9.786°)
Using a calculator, we find the acceleration a is approximately 1.346 m/s^2.
Now, set the final velocity, vf, to zero, and solve for the distance traveled, s:
0 = (3.48 m/s)^2 + 2 * 1.346 m/s^2 * s
Simplifying the equation, we get:
0 = 3.48^2 + 2 * 1.346 * s
Rearranging the equation, we find:
2 * 1.346 * s = - 3.48^2
Solving for s:
s = (-3.48^2) / (2 * 1.346)
Using a calculator, we find that s is approximately -5.45 meters.
Since distance cannot be negative, we take the absolute value of s:
s ≈ 5.45 meters
Therefore, the box travels approximately 5.45 meters along the incline before coming to rest.