A box is sliding up an incline that makes an angle of 16.0° 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 1.30m/s. How far does the box travel along the incline before coming to rest?
To find how far the box travels along the incline before coming to rest, we need to analyze the forces acting on the box.
First, let's resolve the forces acting parallel and perpendicular to the incline:
1. Force parallel to the incline:
- Component of the weight force parallel to the incline: m * g * sin(theta)
- Force due to kinetic friction: coefficient of kinetic friction * Normal force
- In this case, the Normal force is equal to the component of the weight force perpendicular to the incline, which is m * g * cos(theta)
2. Force perpendicular to the incline:
- Component of the weight force perpendicular to the incline: m * g * cos(theta)
Now, we can calculate the force parallel to the incline that acts on the box:
Force parallel to the incline = Component of the weight force parallel to the incline - Force due to kinetic friction
= (m * g * sin(theta)) - (coefficient of kinetic friction * m * g * cos(theta))
= m * g * (sin(theta) - coefficient of kinetic friction * cos(theta))
Since the box comes to rest, we know that the net force acting parallel to the incline is zero:
Force parallel to the incline = 0
Therefore, we can solve for the distance traveled along the incline.
Setting the force equation equal to zero:
0 = m * g * (sin(theta) - coefficient of kinetic friction * cos(theta))
Rearranging the equation to solve for the distance traveled:
sin(theta) = coefficient of kinetic friction * cos(theta)
tan(theta) = coefficient of kinetic friction
theta = arctan(coefficient of kinetic friction)
Now, we can substitute the given values into the equation:
theta = arctan(0.180) ≈ 10.37°
We know that the initial speed of the box at the bottom of the incline is 1.30 m/s.
Using the equation of motion for uniformly accelerated linear motion:
v_f^2 = v_i^2 + 2 * a * d
where
v_f = final velocity (0 m/s because the box comes to rest)
v_i = initial velocity (1.30 m/s)
a = acceleration (g * (sin(theta) - coefficient of kinetic friction * cos(theta)))
d = distance traveled along the incline (what we want to find)
Rearranging the equation to solve for d:
0 = (1.30 m/s)^2 + 2 * (g * (sin(theta) - coefficient of kinetic friction * cos(theta))) * d
Simplifying the equation:
d = -v_i^2 / (2 * a)
Substituting the given values:
d = -(1.30 m/s)^2 / (2 * g * (sin(theta) - coefficient of kinetic friction * cos(theta)))
Now, we can substitute the values of g (acceleration due to gravity) and theta (angle of the incline):
d = -(1.30 m/s)^2 / (2 * 9.8 m/s^2 * (sin(16.0°) - 0.180 * cos(16.0°)))
Evaluating the expression:
d ≈ -0.0945 m
Since distance cannot be negative, we take the absolute value:
d ≈ 0.0945 m
Therefore, the box travels approximately 0.0945 meters along the incline before coming to rest.
Assistance needed.
Please type your subject in the School Subject box. Any other words, including obscure abbreviations, are likely to delay responses from a teacher who knows that subject well.