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 use Newton's second law of motion and the concept of work and energy.
First, let's calculate the gravitational force (Fg) acting on the box. The gravitational force can be calculated using the formula:
Fg = m * g
Where:
m = mass of the box
g = acceleration due to gravity (approximately 9.8 m/s²)
Next, we need to calculate the component of the gravitational force parallel to the incline (Fpar) and the component perpendicular to the incline (Fperp).
Fpar = Fg * sin(θ)
Fperp = Fg * cos(θ)
Where:
θ = angle of the incline (21.8°)
Now, let's calculate the net force (Fnet) acting on the box along the incline. The net force can be calculated as:
Fnet = Fpar - frictional force
The frictional force (Ffriction) can be calculated using the formula:
Ffriction = μ * Fperp
Where:
μ = coefficient of kinetic friction (0.170)
Now we need to calculate the work done by the net force on the box. The work done can be calculated using the formula:
Work = Force * Distance * cos(180°)
Since the work done by the net force equals the change in kinetic energy, we can write:
Work = ΔKE
The change in kinetic energy can be calculated as:
ΔKE = 0.5 * m * (vf² - vi²)
Where:
vi = initial velocity (3.48 m/s)
vf = final velocity (0 m/s)
Since the box comes to rest, the final velocity (vf) is 0 m/s.
Finally, we can substitute the formulas and solve for the distance the box travels along the incline (Distance).
ΔKE = Work
0.5 * m * (0 - vi²) = Fnet * Distance * cos(180°)
Rearranging the formula, we can solve for Distance:
Distance = - (0.5 * m * (0 - vi²)) / (Fnet * cos(180°))
Substituting all the known values, we can find the distance the box travels along the incline.