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EXAMPLE:
A horizontal rope is tied to a 58.0kg box on frictionless ice. What is the tension in the rope if:
1)The box is at rest? – Sol: T=0 N
2)The box moves at a steady vx = 6.00m/s ? – Sol: T=0 N
3)The box vx= 6.00m/s and ax= 5.60m/s^2 ? – Sol: T=325N
QUESTION:
A horizontal rope is tied to a 49.0kg box on frictionless ice. What is the tension in the rope if:
1)The box is at rest? – Sol: T=0 N
2)The box moves at a steady vx= 4.20m/s ? – Sol: T=0 N
3)The box vx= 4.20m/s and ax= 5.60m/s^2 ?
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T = mg + ma T = tension, N, kg-m/s2 m = mass, kg g = gravitational force, 9.8 m/s2 a = acceleration, m/s2♦ Relevant knowledgeThe tension formula:
This problem was solved using Newton second of Motion concepts.
Write the Newton second of motion expression first, then calculate the tension on your rope.
Newton’s second law of motion states that the magnitude and mass of an object that is subject to net force will directly affect its acceleration.
Expression of tension or force.
[katex]T = ma[/katex]
(3) Calculate the tension of the rope.
The expression for tension is: [katex]T = ma[/katex]
Substitute [katex]49.0{\rm{ kg}}[/katex]
Ans of Part 3: [katex]\begin{array}{c}\\T = \left( {49.0{\rm{ kg}}} \right)\left( {5.6{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 274.4{\rm{ N}}\\\end{array}[/katex]
The tension of the rope is [katex]274{\rm{ N}}[/katex]