Eric Q. answered • 03/23/23
Physics Tutor with 5+ years of Teaching Experience
This question is best answered by first drawing a free-body diagram of the box that is of interest, where the applied force is pointing to the right, the force of friction pointing to the left, the weight of the block pointing up, and the normal force pointing up (opposite direction of the weight arrow). After drawing arrows in the direction of the force relative to the block, it would be helpful to write a net force equation to describe the block at the time of applied force (12N).
Fapplied - Ff = ma, a is the acceleration vector.
Knowing that Ff = FN µ, where µ is the coefficient of friction (this coefficient can be either kinetic or static friction, depending on what is relevant in the system.), FN being the normal force exerted by the surface that the block is on.
Since the block is not accelerating in the y-direction, the acceleration in the y-direction is 0. This allows us to say FN = mg.
We can then use this to solve for the net acceleration of the block on the rough surface using two of the equations that we have written out.
Ff = FN µ
Fapplied - Ff = ma
We can substitute FN for mg, subsequently Ff for mgµ.
Fapplied - mgµ = ma
From there, we can solve for the acceleration in terms of the known terms, leading us to the expression:
a = (Fapplied - mgµ)/m
This allows us to substitute all the known variables for the numbers that are given in the problem. Assuming g = 10, the acceleration of the block is a = 2 m/s2.
With our calculated acceleration, we can use kinematics to obtain the final velocity of the block.
vf2= v02 + 2aΔx
Since we know the distance traveled, the acceleration of the block, and the initial velocity of the block, we can substitute the variables in the equation to obtain the final velocity of the block.
After calculation, the final velocity of the block is vf = 6 m/s.
