A 5-kg box is released on a 30 degree incline and accelerates down the incline at 2.79 m/s^2. To the nearest tenth of a Newton, what is the friction force impeding its motion?
16.05 N
To find the friction force impeding the motion of the box, we need to understand the forces acting on the box.
Let's break down the forces first:
1. Gravitational force (mg): The weight of the box is given by the equation F = mg, where m is the mass and g is the acceleration due to gravity (9.8 m/s^2).
Given that the mass of the box is 5 kg, the weight is calculated as follows:
Weight (W) = mass (m) * acceleration due to gravity (g)
W = 5 kg * 9.8 m/s^2 = 49 N
2. Normal force (N): This force is the perpendicular force applied by the incline surface on the box. It is equal to the component of the gravitational force perpendicular to the surface.
Considering the inclination of the surface, the normal force would be given by:
N = mg * cos(θ), where θ is the angle of inclination.
Given that the angle of inclination is 30 degrees, we can calculate the normal force as follows:
N = 5 kg * 9.8 m/s^2 * cos(30 degrees)
Now we can calculate the frictional force:
Frictional force (Ff) = coefficient of friction (μ) * normal force (N)
Since we are not provided with the coefficient of friction, we cannot directly calculate the frictional force.
Therefore, we need additional information to find the coefficient of friction, such as the material of the box and the surface it is sliding on.