A 2.0-m-long, 500 g rope pulls a 10 kg block of ice across a horizontal, frictionless surface. The block accelerates at 2.0 m/s2. How much force pulls forward on he rope? Assume that the rope is perfectly horizontal.

Ch 07: Newton's Third Law

Knight Calc - Physics for Scientists and Engineers 5th Edition

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Knight Calc 5th Edition

Ch 07: Newton's Third Law

Problem 18b

Chapter 7, Problem 18b

A 2.0-m-long, 500 g rope pulls a 10 kg block of ice across a horizontal, frictionless surface. The block accelerates at 2.0 m/s². How much force pulls forward on he rope? Assume that the rope is perfectly horizontal.

Verified step by step guidance

Step 1: Identify the forces acting on the system. The rope pulls the block of ice forward, and the block accelerates due to the tension in the rope. Since the surface is frictionless, there is no opposing force.

Step 2: Use Newton's second law of motion, $ F = ma $, to calculate the net force acting on the block. Here, $ m $ is the mass of the block (10 kg) and $ a $ is its acceleration (2.0 m/s^2). Substitute these values into the formula to find the net force exerted on the block.

Step 3: Recognize that the force pulling forward on the rope is equal to the tension in the rope. The tension in the rope is the same as the force exerted on the block because the rope is perfectly horizontal and there are no other forces acting on it.

Step 4: Consider the mass of the rope (500 g or 0.5 kg). Since the rope is being pulled forward and is part of the system, its mass does not affect the tension calculation directly in this scenario because the acceleration of the rope matches the acceleration of the block.

Step 5: Conclude that the force pulling forward on the rope is equal to the net force calculated in Step 2. This is because the rope transmits the force to the block, and the system is in equilibrium horizontally.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Newton's Second Law of Motion

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This relationship is expressed by the formula F = ma, where F is the net force, m is the mass, and a is the acceleration. In this scenario, understanding this law is crucial to calculate the force exerted on the block of ice and subsequently on the rope.

Tension in a Rope

Tension is the force transmitted through a rope or string when it is pulled tight by forces acting from opposite ends. In this problem, the tension in the rope is responsible for pulling the block of ice. Since the surface is frictionless, the tension must equal the force required to accelerate the block, allowing us to determine the force exerted on the rope.

Frictionless Surface

A frictionless surface is an idealized concept where no frictional forces oppose the motion of objects. This means that all the force applied to the block of ice is used to accelerate it, without any energy lost to friction. Understanding this condition simplifies the analysis, as it allows us to focus solely on the forces acting on the block and the rope without considering frictional effects.

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Textbook Question

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The sled dog in FIGURE EX7.15 drags sleds A and B across the snow. The coefficient of friction between the sleds and the snow is 0.10. If the tension in rope 1 is 150 N, what is the tension in rope 2?

Textbook Question

FIGURE EX7.17 shows two 1.0 kg blocks connected by a rope. A second rope hangs beneath the lower block. Both ropes have a mass of 250 g. The entire assembly is accelerated upward at 3.0 m/s² by force F. What is the tension at the top end of rope 1?

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Textbook Question

The 1.0 kg block in FIGURE EX7.23 is tied to the wall with a rope. It sits on top of the 2.0 kg block. The lower block is pulled to the right with a tension force of 20 N. The coefficient of kinetic friction at both the lower and upper surfaces of the 2.0 kg block is μ_k = 0.40. What is the tension in the rope attached to the wall?

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