# A 2.0-kg block of wood is on a level surface where μs and μk = 0.60. A 13.7-N force is being applied to the block parallel to the surface.(a) If the block was originally at rest, then(A) it will remain at rest, and the force of friction will be about 15.7 N.(B) it will remain at rest, and the force of friction will be about 13.7 N.(C) it will remain at rest, and the force of friction will be 11.8 N.(D) it will begin to slide with a net force of about 1.9 acting on the block.(b) If the block was originally in motion, and the 13.7-N applied force is in the direction of motion, then(A) it will accelerate under a net force of about 1.9 N.(B) it will move at constant speed.(C) it will decelerate under a net force of about 1.9 N.(D) it will decelerate under a net force of 11.8 N.

Deepak Patra
askIITians Faculty 471 Points
9 years ago
The correct option is:

One should consider the effect of weight of the rope at various positions to account for the tension locally.
At the top, the magnitude of tension would be equal to the magnitude of weight because the entire mass of the rope lies below it, and can be given as:

Therefore the magnitude of tension at the top is 0.98 N.
As you traverse down the rope, the effective mass of the rope which lies in the rope below is reduced; therefore the magnitude of the tension keeps on decreasing. One must understand that the tension in the rope appear because the gravity acts on the rope, and the magnitude of this action depends on the mass of the rope below the point (on a rope) in consideration.
At the bottom, there is no weight below to create tension in the string, and therefore this point is free from any tension.
Thus, (d) is the correct option while the others are ruled out.