Ch 05: Applying Newton's Laws

Young & Freedman Calc - University Physics 15th Edition

Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook

Young & Freedman Calc 15th Edition

Ch 05: Applying Newton's Laws

Problem 17b

Chapter 5, Problem 17b

A light rope is attached to a block with mass $4.00$ kg that rests on a frictionless, horizontal surface. The horizontal rope passes over a frictionless, massless pulley, and a block with mass $m$ is suspended from the other end. When the blocks are released, the tension in the rope is $15.0$ N. What is the acceleration of either block?

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Identify the forces acting on each block. For the 4.00 kg block on the horizontal surface, the forces are the tension in the rope (T = 15.0 N) and the normal force balancing its weight. For the hanging block with mass m, the forces are the tension in the rope (T = 15.0 N) acting upward and the gravitational force (m * g) acting downward.

Write Newton's second law for the 4.00 kg block on the horizontal surface. Since it is accelerating to the right, the net force is equal to the tension in the rope: F_net = T = m₁ * a, where m₁ = 4.00 kg and a is the acceleration.

Write Newton's second law for the hanging block. The net force is the difference between the gravitational force and the tension: F_net = m * g - T = m₂ * a, where m₂ is the mass of the hanging block, g is the acceleration due to gravity (9.8 m/s²), and a is the acceleration.

Combine the two equations from steps 2 and 3. Solve for the acceleration (a) by eliminating the tension (T). From step 2: a = T / m₁. Substitute this into the equation from step 3: m * g - T = m₂ * (T / m₁). Rearrange to isolate a.

Simplify the resulting equation to find the acceleration (a) in terms of the given values (T = 15.0 N, m₁ = 4.00 kg, g = 9.8 m/s²). This will give you the acceleration of both blocks, as they are connected by the rope and must have the same magnitude of acceleration.

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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, the tension in the rope and the weight of the suspended block create the net force that determines the acceleration of both blocks.

Tension in a Rope

Tension is the force transmitted through a rope or string when it is pulled tight by forces acting at either end. In this problem, the tension in the rope affects both the block on the frictionless surface and the suspended block. Understanding how tension works is crucial for analyzing the forces acting on each block and calculating their acceleration.

Free Body Diagram

A Free Body Diagram (FBD) is a graphical representation used to visualize the forces acting on an object. By isolating the blocks and drawing FBDs, one can identify all the forces, including tension and gravitational force, acting on each block. This helps in applying Newton's laws to solve for unknown quantities, such as acceleration in this case.

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Free-Body Diagrams

Related Practice

Textbook Question

An $8.00$ -kg block of ice, released from rest at the top of a $1.50$ -m-long frictionless ramp, slides downhill, reaching a speed of $2.50$ m/s at the bottom. What is the angle between the ramp and the horizontal?

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Textbook Question

An $8.00$ -kg block of ice, released from rest at the top of a $1.50$ -m-long frictionless ramp, slides downhill, reaching a speed of $2.50$ m/s at the bottom. What would be the speed of the ice at the bottom if the motion were opposed by a constant friction force of $10.0$ N parallel to the surface of the ramp?

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Textbook Question

A light rope is attached to a block with mass $4.00$ kg that rests on a frictionless, horizontal surface. The horizontal rope passes over a frictionless, massless pulley, and a block with mass m is suspended from the other end. When the blocks are released, the tension in the rope is $15.0$ N. Find $m$ .

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Textbook Question

When jumping straight up from a crouched position, an average person can reach a maximum height of about $60$ cm. During the jump, the person's body from the knees up typically rises a distance of around $50$ cm. To keep the calculations simple and yet get a reasonable result, assume that the entire body rises this much during the jump. With what initial speed does the person leave the ground to reach a height of $60$ cm?

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Textbook Question

A light rope is attached to a block with mass $4.00$ kg that rests on a frictionless, horizontal surface. The horizontal rope passes over a frictionless, massless pulley, and a block with mass $m$ is suspended from the other end. When the blocks are released, the tension in the rope is $15.0$ N. Draw two free-body diagrams: one for each block.

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Textbook Question

A light rope is attached to a block with mass $4.00$ kg that rests on a frictionless, horizontal surface. The horizontal rope passes over a frictionless, massless pulley, and a block with mass $m$ is suspended from the other end. When the blocks are released, the tension in the rope is $15.0$ N. How does the tension compare to the weight of the hanging block?

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