A 5 kg block is being pulled to the right by a rope which is attached to a dangling 2 kg mass. The coefficient of kinetic friction between the mass & table is 0.1 What is the tension in the rope?
To find the tension in the rope, we need to consider the forces acting on the 5 kg block and use Newton's second law of motion (F = ma) to solve for the tension.
First, let's draw a free body diagram to analyze the forces acting on the 5 kg block:
1. The weight of the block acts vertically downward and has a magnitude of (mass of block) x (acceleration due to gravity) = (5 kg) x (9.8 m/s^2) = 49 N.
2. The tension in the rope acts horizontally to the right.
3. The force of kinetic friction opposes the motion and acts horizontally to the left. Its magnitude is given by (coefficient of kinetic friction) x (normal force), where the normal force equals the weight of the block (since the block is on a flat surface with no vertical acceleration). So the force of kinetic friction is (0.1) x (weight of block) = 0.1 x 49 N = 4.9 N.
According to Newton's second law, the net force acting on an object is equal to its mass times its acceleration. In this case, the net force is the sum of the tension and the force of kinetic friction. The acceleration of the block is unknown and will be determined.
49 N (weight of block) - 4.9 N (force of kinetic friction) = mass of block x acceleration
Substituting the given values:
49 N - 4.9 N = 5 kg x acceleration
44.1 N = 5 kg x acceleration
Now we can solve for the acceleration:
acceleration = 44.1 N / 5 kg = 8.82 m/s^2
Finally, since the rope is connected to a 2 kg mass that is hanging, we can use Newton's second law again to find the tension in the rope:
Tension in the rope = weight of hanging mass + (mass of hanging mass x acceleration)
Weight of hanging mass = (mass of hanging mass) x (acceleration due to gravity) = (2 kg) x (9.8 m/s^2) = 19.6 N
Tension in the rope = 19.6 N + (2 kg) x (8.82 m/s^2) = 19.6 N + 17.64 N = 37.24 N
Therefore, the tension in the rope is 37.24 N.