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 17c

Chapter 5, Problem 17c

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 $m$ .

Verified step by step guidance

Step 1: Begin by analyzing the forces acting on the 4.00 kg block on the horizontal surface. Since the surface is frictionless, the only horizontal force acting on this block is the tension in the rope, T = 15.0 N. Using Newton's second law, the acceleration of the block can be expressed as:

a = \(\frac{T}{m_1}\)

, where

m_1 = 4.00 \ \(\text{kg}\)

Step 2: Next, consider the forces acting on the hanging block with mass m. The forces are the gravitational force (

F_g = m \(\cdot\) g

) acting downward and the tension in the rope (T = 15.0 N) acting upward. Using Newton's second law for the hanging block, the net force is:

F_{\(\text{net}\)} = m \(\cdot\) g - T

, and this net force equals the mass of the block times its acceleration:

F_{\(\text{net}\)} = m \(\cdot\) a

Step 3: Since the two blocks are connected by the rope, they share the same magnitude of acceleration, a. Substitute the expression for a from Step 1 into the equation for the hanging block:

m \(\cdot\) \(\frac{T}{m_1}\) = m \(\cdot\) g - T

Step 4: Rearrange the equation from Step 3 to isolate m. Start by expanding and simplifying:

m \(\cdot\) \(\frac{T}{m_1}\) + T = m \(\cdot\) g

. Then, factor out m on the left-hand side:

m \(\cdot\) \(\left\)(\(\frac{T}{m_1}\) + g\(\right\)) = T

. Finally, solve for m:

m = \(\frac{T}{\frac{T}{m_1}\) + g}

Step 5: Substitute the known values into the equation for m. Use

T = 15.0 \ \(\text{N}\)

m_1 = 4.00 \ \(\text{kg}\)

, and

g = 9.80 \ \(\text{m/s}\)^2

. This will give the value of m. Perform the arithmetic to find the final result.

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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

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 law can be expressed with the formula F = ma, where F is the net force, m is the mass, and a is the acceleration. In this scenario, it helps to determine the relationship between the forces acting on both blocks and how they influence their motion.

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 affects both the block on the surface and the suspended block. Understanding how tension works is crucial for calculating the mass of the suspended block, as it directly influences the forces acting on both blocks.

Free Body Diagram

A Free Body Diagram (FBD) is a graphical representation used to visualize the forces acting on an object. By isolating the block on the surface and the suspended block, we can identify all the forces, including tension and gravitational force. Analyzing these forces through FBDs allows for a clearer understanding of the system's dynamics and aids in solving for unknown quantities, such as the mass m.

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Related Practice

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. What is the acceleration of either block?

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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

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. Draw a free-body diagram of the person during the jump.

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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?