/*! 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 30 A small block has constant accel... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(6.80 \mathrm{~m}\) to the bottom of the incline is \(3.80 \mathrm{~m} / \mathrm{s}\). What is the speed of the block when it is \(3.40 \mathrm{~m}\) from the top of the incline?

Short Answer

Expert verified
The speed of the block when it is \(3.40 \mathrm{~m}\) from the top of the incline is approximately \(2.69 \mathrm{m/s}\).

Step by step solution

01

Understand available data and find acceleration

The block is moving from rest, so the initial velocity (U) is 0. The final velocity (V) when the block traveled \(6.80 \mathrm{~m}\) is \(3.80 \mathrm{~m} / \mathrm{s}\). The distance (s) covered is \(6.80 \mathrm{~m}\). The acceleration (a) can be found using the equation \( v^2 = u^2 + 2as \), thus the acceleration \(a = (v^2 - u^2) / (2s)\). Substituting values, we find \(a = (3.80^2 - 0^2) / (2*6.80) \approx 1.06 \mathrm{m/s^2}\).
02

Use the acceleration to find the velocity at 3.40m

Now we want to find the velocity when the block has traveled \(3.40 \mathrm{~m}\). We already have the value of acceleration, and the distance is given. We can calculate the velocity using the equation \( v^2 = u^2 + 2as \). Substituting values, we find \(v = sqrt(0^2 + 2*1.06*3.40)\). Therefore, \(v \approx 2.69 \mathrm{m/s}\).

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Key Concepts

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

Constant Acceleration
When we talk about constant acceleration in physics, we refer to an object that's changing its velocity at a steady rate over time—imagine a car increasing its speed by the same amount every second. This is a crucial concept when dealing with kinematics, the study of motion without considering the causes.

In our exercise, the block sliding down the frictionless incline has constant acceleration, which means it picks up speed uniformly as it descends. Understanding constant acceleration allows us to predict the future motion of the block, such as calculating how fast it will be moving at a certain point down the incline.
Kinematic Equations
Kinematic equations are the formulae that describe the motion of objects under constant acceleration without accounting for the forces that cause this motion. They are incredibly useful for solving problems in physics where the motion is predictable and follows certain principles. A key kinematic equation that helps us solve our textbook problem is: \( v^2 = u^2 + 2as \), where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(s\) is the displacement of the object.

By using this equation, we were able to determine the block's speed at various points on the incline, provided we know its initial speed, the acceleration, and the distance traveled.
Frictionless Incline Motion
The problem we're tackling involves a block moving down a frictionless incline. This is an idealized scenario in physics where an incline's surface does not exert any resistive force against the block's motion—the block doesn't slow down due to friction as it slides. The absence of friction allows us to simplify our calculations since we only consider the force of gravity component along the incline and the acceleration it causes.

This simplification directly leads us to the concept of constant acceleration, as the object in question will accelerate at a consistent rate down the slope, making it a textbook example for applying our kinematic equations.
Initial Velocity
Initial velocity is the speed at which an object starts its journey. It's an important starting parameter for solving motion-related problems in physics. In our exercise, the block starts from rest, which means its initial velocity is zero. This simplifies our calculations, as it allows us to predict the future velocities and positions of the object using kinematic equations with one less variable to worry about.

Knowing the initial velocity is crucial, as all other kinematic values depend on this starting point. In scenarios like this, where the object is released from rest, many kinematic equations become more intuitive and manageable.

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Most popular questions from this chapter

A car and a truck start from rest at the same instant, with the car initially at some distance behind the truck. The truck has a constant acceleration of \(2.10 \mathrm{~m} / \mathrm{s}^{2},\) and the car has an acceleration of \(3.40 \mathrm{~m} / \mathrm{s}^{2}\). The car overtakes the truck after the truck has moved \(60.0 \mathrm{~m}\). (a) How much time does it take the car to overtake the truck? (b) How far was the car behind the truck initially? (c) What is the speed of each when they are abreast? (d) On a single graph, sketch the position of each vehicle as a function of time. Take \(x=0\) at the initial location of the truck.

You throw a rock straight up and find that it returns to your hand \(3.60 \mathrm{~s}\) after it left your hand. Neglect air resistance. What was th maximum height above your hand that the rock reached?

The acceleration of a bus is given by \(a_{x}(t)=\alpha t,\) where \(\alpha=1.2 \mathrm{~m} / \mathrm{s}^{3} .\) (a) If the bus's velocity at time \(t=1.0 \mathrm{~s}\) is \(5.0 \mathrm{~m} / \mathrm{s}\) what is its velocity at time \(t=2.0 \mathrm{~s} ?\) (b) If the bus's position at time \(t=1.0 \mathrm{~s}\) is \(6.0 \mathrm{~m},\) what is its position at time \(t=2.0 \mathrm{~s} ?(\mathrm{c})\) Sketch \(a_{y}-t\) \(v_{y}-t,\) and \(x-t\) graphs for the motion.

You are standing at rest at a bus stop. A bus moving at a constant speed of \(5.00 \mathrm{~m} / \mathrm{s}\) passes you. When the rear of the bus is \(12.0 \mathrm{~m}\) past you, you realize that it is your bus, so you start to run toward it with a constant acceleration of \(0.960 \mathrm{~m} / \mathrm{s}^{2}\). How far would you have to run before you catch up with the rear of the bus, and how fast must you be running then? Would an average college student be physically able to accomplish this?

An egg is thrown nearly vertically upward from a point near the cornice of a tall building. The egg just misses the cornice on the way down and passes a point \(30.0 \mathrm{~m}\) below its starting point \(5.00 \mathrm{~s}\) after it leaves the thrower's hand. Ignore air resistance. (a) What is the initial speed of the egg? (b) How high does it rise above its starting point? (c) What is the magnitude of its velocity at the highest point? (d) What are the magnitude and direction of its acceleration at the highest point? (e) Sketch \(a_{y}-t, v_{y}-t,\) and \(y-t\) graphs for the motion of the egg.

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