A) Well, well, well, looks like we have a slippery situation here! To find the minimum angle at which the box starts to slip, we need to figure out the maximum value of the static friction force. This occurs just before the box starts to slide. So, we set the maximum static friction force equal to the force tending to make the box slide, which is the product of the mass and acceleration due to gravity, mg, multiplied by the angle, α.
Now, the equation for static friction is Fs = μs*N, where Fs is the static friction force, μs is the coefficient of static friction, and N is the normal force. The normal force is equal to mg*cos(α) since the box is on a ramp. So, we have Fs = μs*(mg*cos(α)), where Fs can have a maximum value of μs*m*g. So, setting these equal, we get:
μs*(mg*cos(α)) = μs*m*g
cos(α) = 1
α = arccos(1)
α = 0 degrees.
Woah, seems like even a super tiny angle is enough for the box to start slipping! Watch out for those slippery textbooks!
B) Once the box has begun to move, it experiences kinetic friction. The equation for kinetic friction is Fk = μk*N, where Fk is the kinetic friction force and μk is the coefficient of kinetic friction. Now, the normal force N is equal to mg*cos(α). The net force acting on the box is the component of the gravitational force perpendicular to the ramp, which is mg*sin(α), minus the kinetic friction force. So, we have:
Net Force = mg*sin(α) - μk*(mg*cos(α))
Using Newton's second law, F = ma, where F is the net force and m is the mass, we can find the acceleration a:
mg*sin(α) - μk*(mg*cos(α)) = m*a
a = g*(sin(α) - μk*cos(α))
Now plug in the values of g (acceleration due to gravity) and μk (coefficient of kinetic friction) to find the magnitude of acceleration.
C) To find the final speed of the box after it has slid a distance of 4.8 m along the loading ramp, we can use the kinematic equation:
vf^2 = vi^2 + 2*a*d
Here, vi is the initial velocity of the box, which is 0 m/s since it starts from rest, a is the acceleration we found in part B, and d is the distance traveled, which is 4.8 m. Plug in these values and solve for vf to find the final velocity.