/*! 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 28 A block of ice with mass 2.00 \(... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A block of ice with mass 2.00 \(\mathrm{kg}\) slides 0.750 \(\mathrm{m}\) down an inclined plane that slopes downward at an angle of \(36.9^{\circ}\) below the horizontal. If the block of ice starts from rest, what is its final speed? You can ignore friction.

Short Answer

Expert verified
The final speed of the block is approximately 2.97 m/s.

Step by step solution

01

Identify the Forces

We start by identifying the forces acting on the block. The only force causing the block to slide down is the component of gravitational force parallel to the inclined plane. This force is given by \( F = mg \sin \theta \), where \( m \) is the mass of the block (2.00 kg), \( g \) is the acceleration due to gravity (9.81 m/s²), and \( \theta \) is the angle of the incline (36.9°).
02

Apply Newton's Second Law

By applying Newton's second law in the direction of the incline, we have \( F = ma \). Thus, the acceleration \( a \) can be calculated as \( a = g \sin \theta \). Substituting \( g = 9.81 \text{ m/s}^2 \) and \( \theta = 36.9^ ext{o} \), we find \( a = 9.81 \times \sin(36.9^ ext{o}) \approx 5.88 \text{ m/s}^2 \).
03

Use Kinematic Equations

To find the final speed, we can use the kinematic equation \( v^2 = u^2 + 2as \), where \( u \) is the initial speed (0 m/s, since it starts from rest), \( a \) is the acceleration (5.88 m/s²), and \( s \) is the distance down the plane (0.750 m).
04

Calculate the Final Speed

Substitute the known values into the kinematic equation: \( v^2 = 0 + 2 \times 5.88 \times 0.750 \). Simplifying gives \( v^2 = 8.82 \). Taking the square root of both sides gives the final speed \( v \approx 2.97 \text{ m/s} \).

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 91Ó°ÊÓ!

Key Concepts

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

Inclined Plane
An inclined plane is a flat surface tilted at an angle to the horizontal. It's a type of simple machine that allows an object to be moved across a distance with less effort. In physics problems, the inclined plane helps us explore the effect of gravitational force on objects that are sliding or rolling down a slope.
When you're working with inclined planes, you need to consider components of forces. The gravitational force acting on an object can be resolved into two components:
  • Parallel to the incline: This component causes the object to slide down.
  • Perpendicular to the incline: This component presses the object against the surface.
On a frictionless inclined plane, as in this problem, only the parallel component influences the motion. Understanding how forces decompose on an inclined plane allows us to analyze movement more efficiently.
Newton's Second Law
Newton's second law of motion is the foundation for understanding how forces affect motion. It states that the acceleration (\(a\)) of an object is directly proportional to the net force (\(F\)) acting on it and inversely proportional to its mass (\(m\)).
The formula can be expressed as: \[ F = ma \]In the context of the inclined plane problem, the net force causing the block to accelerate down the slope is the component of its weight parallel to the incline: \( F = mg\sin\theta \). Using this expression, we can determine the block's acceleration. Solving problems with inclined planes requires applying Newton's second law to each component of the forces involved.
Gravitational Force
Gravitational force is the force of attraction between two masses. On Earth, this force gives weight to physical objects and causes them to fall when dropped. Its standard value is approximately \(9.81\text{ m/s}^2\).
In problems involving inclined planes, gravitational force must be split into components:
  • The component parallel to the plane (\(mg\sin\theta \)) propels the object down the slope.
  • The perpendicular component (\(mg\cos\theta \)) presses the object against the plane's surface.
By evaluating these components, we can better understand how gravity affects the object's movement along the inclined plane.
Kinematic Equations
Kinematic equations describe motion with variables like velocity, acceleration, and displacement. These equations are vital in solving motion-related physics problems.
For the block of ice moving down the inclined plane, we use the equation: \[ v^2 = u^2 + 2as \]Here, \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(s\) is the displacement.
In this exercise, the block starts from rest so \(u = 0\). Plugging in the values, including the calculated acceleration from Newton's second law, allows us to determine the final speed of the block. Understanding how to effectively use kinematic equations is essential for accurately predicting the motion of objects.

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 20.0 -kg rock is sliding on a rough, horizontal surface at 8.00 \(\mathrm{m} / \mathrm{s}\) and eventually stops due to friction. The coefficient of kinetic friction between the rock and the surface is \(0.200 .\) What average power is produced by friction as the rock stops?

Use the work-energy theorem to solve each of these problems. You can use Newton's laws to check your answers. Neglect air resistance in all cases. (a) A branch falls from the top of a 95.0 -m-tall redwood tree, starting from rest. How fast is it moving when it reaches the ground? (b) A volcano ejects a boulder directly upward 525 \(\mathrm{m}\) into the air. How fast was the boulder moving just as it left the volcano? (c) A skier moving at 5.00 \(\mathrm{m} / \mathrm{s}\) encounters a long, rough horizontal patch of snow having coefficient of kinetic friction 0.220 with her skis. How far does she travel on this patch before stopping? (d) Suppose the rough patch in part (c) was only 2.90 m long? How fast would the skier be moving when she reached the end of the patch? (e) At the base of a frictionless icy hill that rises at \(25.0^{\circ}\) above the horizontal, a toboggan has a speed of 12.0 \(\mathrm{m} / \mathrm{s}\) toward the hill. How high vertically above the base will it go before stopping?

You are asked to design spring bumpers for the walls of a parking garage. A freely rolling \(1200-\mathrm{kg}\) car moving at 0.65 \(\mathrm{m} / \mathrm{s}\) is to compress the spring no more than 0.090 \(\mathrm{m}\) before stopping. What should be the force constant of the spring? Assume that the spring has negligible mass.

Magnetar. On December \(27,2004,\) astronomers observed the greatest flash of light ever recorded from outside the solar system. It came from the highly magnetic neutron star SGR \(1806-20\) (a magnetar). During 0.20 \(\mathrm{s}\) , this star released as much energy as our sun does in \(250,000\) years. If \(P\) is the average power output of our sun, what was the average power output (in terms of \(P )\) of this magnetar?

Leg Presses. As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arranged side by side so that they are parallel to each other. When you push the platform, you compress the springs. You do 80.0 J of work when you compress the springs 0.200 \(\mathrm{m}\) from their uncompressed length.(a) What magnitude of force must you apply to hold the platform in this position? (b) How much additional work must you do to move the platform 0.200 \(\mathrm{m}\) farther, and what maximum force must you apply?

See all solutions

Recommended explanations on Physics Textbooks

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.