/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 57 A block of mass \(m_{1}=3.70 \ma... [FREE SOLUTION] | 91Ó°ÊÓ

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A block of mass \(m_{1}=3.70 \mathrm{~kg}\) on a frictionless plane inclined at angle \(\theta=30.0^{\circ}\) is connected by a cord over a massless, frictionless pulley to a second block of mass \(m_{2}=2.30 \mathrm{~kg}\) (Fig. 5-52). What are (a) the magnitude of the acceleration of each block, (b) the direction of the acceleration of the hanging block, and (c) the tension in the cord?

Short Answer

Expert verified
(a) Acceleration = 0.835 m/s², (b) Downward for m₂, (c) Tension = 20.8 N.

Step by step solution

01

Analyze the Forces for Each Block

For the block on the incline (mass = \(m_1 = 3.70 \, \text{kg}\)):- The gravitational force component along the incline is \(m_1 g \sin(\theta)\).For the hanging block (mass = \(m_2 = 2.30 \, \text{kg}\)):- The gravitational force is \(m_2 g\).Both blocks have the same magnitude of acceleration \(a\) because they are connected by the cord.
02

Write the Equations of Motion

For the block on the incline \(m_1\):\[T - m_1 g \sin(\theta) = m_1 a\]For the hanging block \(m_2\):\[m_2 g - T = m_2 a\]
03

Solve for Acceleration

Add the two equations from Step 2 to eliminate the tension \(T\):\[T - m_1 g \sin(\theta) + m_2 g - T = m_1 a + m_2 a\]\[m_2 g - m_1 g \sin(\theta) = (m_1 + m_2) a\]Solve for \(a\):\[a = \frac{m_2 g - m_1 g \sin(\theta)}{m_1 + m_2}\]Plug in the values:\[a = \frac{2.30 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 - 3.70 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \cdot \sin 30.0^\circ}{3.70 \, \text{kg} + 2.30 \, \text{kg}} \approx 0.835 \, \text{m/s}^2\]
04

Determine the Direction of Acceleration for the Hanging Block

Since \(m_2 g > m_1 g \sin(\theta)\), the gravitational force pulling the hanging block downwards is greater than the component of gravitational force pulling the inclined block upwards. Therefore, the hanging block accelerates downward.
05

Calculate the Tension in the Cord

Using either equation from Step 2, let's use the equation for the hanging block:\[m_2 g - T = m_2 a\]Solve for \(T\):\[T = m_2 g - m_2 a\]Plug in the values:\[T = 2.30 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 - 2.30 \, \text{kg} \cdot 0.835 \, \text{m/s}^2 \approx 20.8 \, \text{N}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Frictionless Plane
In physics, a frictionless plane is an idealized surface on which objects can move without any resistance from friction. Imagine a perfectly smooth surface where no force opposes the movement of an object. This idea helps simplify complex problems, allowing students to focus only on the essential forces acting upon the objects.
In real life, all surfaces have some amount of friction, but in theoretical physics problems, imagining a frictionless scenario helps greatly in understanding fundamental principles.
  • On a frictionless plane, only gravity influences the motion of an object.
  • The normal force acts perpendicular to the surface to balance the component of gravity that acts towards the plane.
  • On inclined planes, friction is usually ignored for simplification, focusing instead on gravitational forces and their components.
In our exercise, the block on this frictionless inclined plane is only affected by gravity, which makes problems simpler and computations easier.
Pulley Systems
A pulley system is a simple machine that helps to change the direction of a force applied from one end of a rope or cord. Pulleys are employed in various applications, from simple household tasks to intricate engineering operations.
In the context of physics, pulleys allow us to comprehend the dynamics of movement better, especially when two objects are connected by a cord around the pulley like in the exercise outlined here.
  • A pulley changes the direction of the tension in the cord, which means if one end is pulled downwards with a force, the opposite end has an equal force pulling upwards.
  • When pulleys are considered massless and frictionless, they do not affect the total energy or forces in a system, simplifying calculations.
In our scenario, using a massless, frictionless pulley allows us to easily connect the block on the inclined plane with the hanging block, focusing only on gravitational forces and tension without the worry of additional complexities.
Inclined Plane Dynamics
Inclined plane dynamics refers to studying how objects behave when they move on sloped surfaces. A critical part of understanding these dynamics is breaking down forces into components parallel and perpendicular to the plane's surface.
On an incline, gravity can be decomposed into two components: one acting down the plane and one acting perpendicular to it. Calculating these components allows us to predict the motion accurately.
  • The force acting downwards along the plane is calculated as: \(m_1 g \sin(\theta)\)
  • The perpendicular force is usually balanced by the normal force the plane exerts.
In our exercise, the block on the inclined plane feels a force pulling it downwards along the slope due to gravity. This force competes with the gravitational force acting on the hanging block, thereby affecting the acceleration of the system.
Tension in Cords
Tension in cords is the force conducted along the rope or string when it is pulled tightly by forces acting from opposite ends. Tension is always directed along the length of the cord, and in the case of massless cords, the tension is uniform throughout.
Understanding tension is crucial when solving problems involving multiple connected objects through strings or cords, as it dictates the movement and forces experienced by the objects.
  • The tension in the cord generates equal and opposite forces on each block.
  • In problems like ours, tension balances out part of the gravitational forces, changing the net forces acting on the blocks and influencing acceleration.
Our exercise calculates the tension by examining the forces each block experiences, particularly focusing on the differences in gravitational forces on the inclined block and the hanging block, resulting in a specific tension that allows for their connected acceleration.

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Most popular questions from this chapter

A block is projected up a frictionless inclined \(\begin{array}{llll}\text { plane } & \text { with } & \text { initial } & \text { speed } & v_{0}=\end{array}\) \(3.50 \mathrm{~m} / \mathrm{s} .\) The angle of incline is \(\theta=32.0^{\circ} .\) (a) How far up the plane does the block go? (b) How long does it take to get there? (c) What is its speed when it gets back to the bottom?

A hot-air balloon of mass \(M\) is descending vertically with downward acceleration of magnitude \(a\). How much mass (ballast) must be thrown out to give the balloon an upward acceleration of magnitude \(a\) ? Assume that the upward force from the air (the lift) does not change because of the decrease in mass.

The high-speed winds around a tornado can drive projectiles into trees, building walls, and even metal traffic signs. In a laboratory simulation, a standard wood toothpick was shot by pneumatic gun into an oak branch. The toothpick's mass was \(0.13 \mathrm{~g}\), its speed before entering the branch was \(220 \mathrm{~m} / \mathrm{s},\) and its penetration depth was \(15 \mathrm{~mm} .\) If its speed was decreased at a uniform rate, what was the magnitude of the force of the branch on the toothpick?

The Zacchini family was renowned for their human-cannonball act in which a family member was shot from a cannon using either elastic bands or compressed air. In one version of the act, Emanuel Zacchini was shot over three Ferris wheels to land in a net at the same height as the open end of the cannon and at a range of \(69 \mathrm{~m} .\) He was propelled inside the barrel for \(5.2 \mathrm{~m}\) and launched at an angle of \(53^{\circ} .\) If his mass was \(85 \mathrm{~kg}\) and he underwent constant acceleration inside the barrel, what was the magnitude of the force propelling him? (Hint: Treat the launch as though it were along a ramp at \(53^{\circ} .\) Neglect air drag.)

A car that weighs \(1.30 \times 10^{4} \mathrm{~N}\) is initially moving at \(40 \mathrm{~km} / \mathrm{h}\) when the brakes are applied and the car is brought to a stop in \(15 \mathrm{~m} .\) Assuming the force that stops the car is constant, find (a) the magnitude of that force and (b) the time required for the change in speed. If the initial speed is doubled, and the car experiences the same force during the braking, by what factors are (c) the stopping distance and (d) the stopping time multiplied? (There could be a lesson here about the danger of driving at high speeds.)

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