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 determine how far the box travels along the incline before coming to rest, we can use the principles of physics and apply Newton's second law, as well as the concepts of work and energy.
Step 1: Start by analyzing the forces acting on the box.
There are two relevant forces: the gravitational force pulling the box down the incline and the kinetic friction force opposing the motion of the box. The gravitational force can be split into two components:
- The component of the gravitational force parallel to the incline, which acts to accelerate the box down the incline.
- The component of the gravitational force perpendicular to the incline, which is canceled out by the normal force of the incline.
The kinetic friction force opposes the motion of the box and acts parallel to the incline.
Step 2: Calculate the net force acting on the box.
The net force can be determined using Newton's second law, which states that the net force is equal to the mass of an object multiplied by its acceleration.
Since the box is slowing down and comes to rest, the net force is zero. Thus, we can set up an equation:
Net force = m * a = 0
Step 3: Express the net force in terms of the forces acting on the box.
The net force is the difference between the gravitational force component parallel to the incline and the kinetic friction force. This can be written as:
Net force = m * g * sin(θ) - F_k
where m is the mass of the box, g is the acceleration due to gravity, θ is the angle of the incline, and F_k is the kinetic friction force.
Step 4: Calculate the kinetic friction force.
The kinetic friction force can be determined using the coefficient of kinetic friction (µ_k) multiplied by the normal force (N) acting on the box:
F_k = µ_k * N
The normal force can be found by multiplying the gravitational force component perpendicular to the incline by cos(θ):
N = m * g * cos(θ)
Step 5: Substitute the expressions for the net force and the kinetic friction force into the net force equation.
m * g * sin(θ) - F_k = 0
m * g * sin(θ) - µ_k * N = 0
Substitute the expression for the normal force (N) into the equation:
m * g * sin(θ) - µ_k * m * g * cos(θ) = 0
Step 6: Solve the equation for the distance (d) traveled by the box.
To find the distance traveled by the box, we need to determine the work done on the box. The work done can be calculated as the product of the force applied (net force) and the distance traveled (d):
Work = Force * Distance
The force applied is the average of the initial and final net forces acting on the box, which is zero. Thus, the work done is also zero:
Work = (Initial force + Final force) / 2 * d = 0
Since the work done is zero, we can solve for the distance (d):
(d/2) * (m * g * sin(θ) - µ_k * m * g * cos(θ)) = 0
Simplifying the equation:
d * (m * g * sin(θ) - µ_k * m * g * cos(θ)) = 0
Since the coefficient of kinetic friction (µ_k) is greater than zero, the term in parentheses must equal zero:
m * g * sin(θ) - µ_k * m * g * cos(θ) = 0
Finally, solve for the distance traveled (d):
d = µ_k * m * g * cos(θ) / (m * g * sin(θ))
Simplifying the equation:
d = µ_k * cos(θ) / sin(θ)
Step 7: Calculate the distance traveled by plugging in the given values.
d = 0.170 * cos(21.8°) / sin(21.8°)
d ≈ 0.170 * 0.935 / 0.375
d ≈ 0.170 * 2.493
d ≈ 0.423 m
Therefore, the box travels approximately 0.423 meters along the incline before coming to rest.