A horizontal rope is tied to a 50.0kg box on frictionless ice.

What is the tension in the rope if,

A. the box is at rest

B. the box moves at a steady 5.0m/s

C. the box has v_x=5.0m/s and a_x=5.0m/s^2

Base on the problem about a horizontal rope that is tied to a 50kg box on frictionless ice, calculate the tension rope of the following:

A. the box is at rest

- tension is 0N

B. the box moves at a steady 5.0m/s

-tension is 250N

C. the box has v_x=5.0m/s and a_x=5.0m/s^2

-tension is 250 N.

a) 0

b) 0

c) 250 N

Explanation:

A) When the box is at rest, it has zero acceleration, consequenly, the net force acting on the box is zero, and the rope has no tension. Tension = 0.

B) When the box is moving at a setady 5.0 m/s, again it has zero acceleration. So the same explanation above applies: net force = 0, and tension = 0.

C) When the box has Vx = 5.0 m/s and Ax = 5.0 m/s², you use the second law of Newton to find the net force:

F = mass * acceleration = 50 kg * 5 m/s² = 250 N

Since the box is on frictionless ice, there is not friction force, and the tension is equal to the net force. Tension = 250 N.