/*! 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 33 A small block has constant accel... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 33

A small block has constant acceleration as it slides down a frictionless incline. The block is released from rest at the top of the incline, and its speed after it has traveled 6.80 m to the bottom of the incline is 3.80 m/s. What is the speed of the block when it is 3.40 m from the top of the incline?

Short Answer

Expert verified
The speed of the block at 3.40 m is approximately 2.68 m/s.

Step by step solution

01

Understand the Problem

We have a block sliding down a frictionless incline. It is released from rest, so its initial velocity is 0. We are given the distance traveled (6.80 m) and the final speed at that distance (3.80 m/s). We need to find the speed when the block is halfway down, at 3.40 m.
02

Use the Kinematic Equation

We can use the kinematic equation: \( v^2 = u^2 + 2as \) to find the acceleration \(a\), where \( v \) is the final velocity, \( u \) is the initial velocity, and \( s \) is the distance.
03

Calculate the Acceleration

Given \( u = 0 \), \( v = 3.80 \text{ m/s} \), and \( s = 6.80 \text{ m}\), substitute these into the equation to find \( a \): \( (3.80)^2 = 0 + 2a(6.80) \). Solve for \( a \).
04

Solve for Acceleration

Calculate \( a \): \[ 14.44 = 13.6a \] \[ a = \frac{14.44}{13.6} \approx 1.062 \text{ m/s}^2 \].
05

Determine Speed at 3.40 m

Now use the same kinematic equation to find the speed at 3.40 m: \( v^2 = 0 + 2(1.062)(3.40) \).
06

Calculate the Speed

Perform the calculations: \( v^2 = 2(1.062)(3.40) \), which simplifies to \( v^2 = 7.2088 \). Thus, \( v = \sqrt{7.2088} \approx 2.68 \text{ m/s} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Frictionless Incline
Imagine you're on a playground slide that is completely smooth, with no bumps, grooves, or any resistance to slow you down 鈥� this is what a frictionless incline is like.
When we say 'frictionless,' it means there's no friction acting against the object sliding down the incline.
Without friction, the only force acting on the block is gravity, pulling it straight down towards the center of the Earth.
  • Friction usually opposes motion, so without it, motion becomes simpler to analyze.
  • Since the incline is frictionless, calculations can assume no energy is lost due to friction.
  • This simplification allows us to focus solely on gravitational forces and acceleration.
Understanding this concept is crucial in kinematics, where we're often asked to analyze the motion of objects without the complicated factors of friction.
We'll use this understanding to predict how objects move down slanted surfaces.
Constant Acceleration
When an object has constant acceleration, its speed increases steadily over time.
In our problem, the block accelerates constantly as it moves down the frictionless incline.
This steady increase in speed makes it easier to use kinematic equations to understand how the block's motion progresses.
  • Constant acceleration means that the gravitational force on the block doesn't change.
  • You can think of constant acceleration as a push that keeps adding the same amount of speed in every second.
  • With constant acceleration, calculations become more predictable and straightforward.
Having constant acceleration simplifies the analysis, as you don't have to account for varying speeds or changing forces.
This enables us to accurately predict how fast the block is moving at different points along the incline.
Kinematic Equations
Kinematic equations help us describe the motion of objects under constant acceleration.
In this context, the kinematic equation we use is:
\( v^2 = u^2 + 2as \), where:
  • \( v \) is the final velocity, or speed, you're trying to find.
  • \( u \) is the initial velocity, or the speed at the start (which is zero if it's released from rest).
  • \( a \) is the acceleration, which we calculated previously.
  • \( s \) is the distance the block has traveled down the incline.
This equation lets us determine what happens to the speed of the block as it travels down.
By substituting known values, we solve for the unknown variable, making predictions about the block's velocity at various points.
Initial Velocity
In our scenario of a block on a frictionless incline, the initial velocity is crucial.
The block is released from rest, meaning its initial velocity, \( u \), is 0 m/s.
This absence of an initial push or speed is what allows us to simplify our calculations using the kinematic equations.
  • Knowing the initial velocity helps us determine how much speed the block will gain as it accelerates.
  • When an object starts from rest, kinematic equations become more straightforward, as they often eliminate terms involving initial velocity.
  • In this case, starting from rest means that all the motion observed is due to the constant acceleration along the incline.
Understanding the starting conditions, such as the initial velocity, is key to all subsequent calculations, leading us to find out how fast the block travels over a given distance.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) If a flea can jump straight up to a height of 0.440 m, what is its initial speed as it leaves the ground? (b) How long is it in the air?

A rocket carrying a satellite is accelerating straight up from the earth's surface. At 1.15 s after liftoff, the rocket clears the top of its launch platform, 63 m above the ground. After an additional 4.75 s, it is 1.00 km above the ground. Calculate the magnitude of the average velocity of the rocket for (a) the 4.75-s part of its flight and (b) the first 5.90 s of its flight.

The human body can survive an acceleration trauma incident (sudden stop) if the magnitude of the acceleration is less than 250 m/s\(^{2}\). If you are in an automobile accident with an initial speed of 105 km/h (65 mi/h) and are stopped by an airbag that inflates from the dashboard, over what distance must the airbag stop you for you to survive the crash?

A typical male sprinter can maintain his maximum acceleration for 2.0 s, and his maximum speed is 10 m/s. After he reaches this maximum speed, his acceleration becomes zero, and then he runs at constant speed. Assume that his acceleration is constant during the first 2.0 s of the race, that he starts from rest, and that he runs in a straight line. (a) How far has the sprinter run when he reaches his maximum speed? (b) What is the magnitude of his average velocity for a race of these lengths: (i) 50.0 m; (ii) 100.0 m; (iii) 200.0 m?

A bird is flying due east. Its distance from a tall building is given by \(x(t) =\) 28.0 m \(+\) (12.4 m/s)\(t\) - (0.0450 m/s\(^3)t^3\). What is the instantaneous velocity of the bird when \(t =\) 8.00 s?
See all solutions

Recommended explanations on Physics Textbooks

Further Mechanics and Thermal Physics

Read Explanation

Medical Physics

Read Explanation

Linear Momentum

Read Explanation

Circular Motion and Gravitation

Read Explanation

Radiation

Read Explanation

Electricity and Magnetism

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.