A 16.0 kg box is released on a 39.0° incline and accelerates down the incline at 0.267 m/s2. Find the friction force impeding its motion
Use F = m a, along the direction of motion. The net force pusing the box downhill is
M g sin39 - Ff = M a
Solve for Ff, the friction force.
Please do not put your grade level as the subject.
To find the friction force impeding the motion of the box, we need to consider the forces acting on it. In this case, the relevant forces are the force of gravity and the force due to the acceleration down the incline.
First, let's find the force of gravity acting on the box. The force of gravity (weight) can be calculated by multiplying the mass of the box (16.0 kg) by the acceleration due to gravity (9.8 m/s^2).
Weight = mass * acceleration due to gravity
Weight = 16.0 kg * 9.8 m/s^2 = 156.8 N
Next, let's break down the weight force into its components parallel and perpendicular to the incline. The weight force can be resolved into two components: the force parallel to the incline, which causes the acceleration down the incline, and the force perpendicular to the incline.
Force parallel to the incline = Weight * sin(θ)
Force perpendicular to the incline = Weight * cos(θ)
where θ is the angle of incline, which is 39.0° in this case.
Force parallel to the incline = 156.8 N * sin(39.0°) = 95.1 N
Force perpendicular to the incline = 156.8 N * cos(39.0°) = 120.6 N
Since the box is accelerating down the incline, the force parallel to the incline is greater than the opposing force of friction. Therefore, the friction force is equal in magnitude and opposite in direction to the parallel component of the weight force.
Friction force = Force parallel to the incline = 95.1 N
So, the friction force impeding the motion of the box is 95.1 N.