/*! 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 22 Draw a free-body diagram of a bl... [FREE SOLUTION] | 91Ó°ÊÓ

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Draw a free-body diagram of a block which slides down a frictionless plane having an inclination of \(\theta=15.0^{\circ}\) (Fig. P5.22). The block starts from rest at the top and the length of the incline is 2.00 \(\mathrm{m}\) . Find (a) the acceleration of the block and \((b)\) its speed when it reaches the bottom of the incline.

Short Answer

Expert verified
The acceleration of the block is \(a = g\sin(15^\circ) = 9.81\sin(15^\circ)\, \mathrm{m/s^2}\) and its speed when it reaches the bottom is calculated using \(v = \sqrt{2as}\).

Step by step solution

01

Free-Body Diagram

Draw a block on an inclined plane at an angle \(\theta = 15.0^\circ\) to the horizontal. Indicate the force due to gravity (weight) acting vertically downward. This force can be resolved into two components: one parallel to the incline (\(mg\sin\theta\)) and one perpendicular to the incline (\(mg\cos\theta\)). There is no friction, so no force is opposing the motion down the plane.
02

Calculate the Acceleration

Use Newton's second law along the incline (\(F = ma\)) to find the acceleration. The only force acting along the incline is the component of gravity parallel to it, or \(mg\sin\theta\). Thus, \(mg\sin\theta = ma\). Canceling mass from both sides and using g = 9.81 \(\mathrm{m/s^2}\), the acceleration a can be calculated as \(a = g\sin\theta\).
03

Calculate Block's Speed at the Bottom

Use kinematic equation \(v^2 = u^2 + 2as\) where u is the initial speed, v is the final speed, a is the acceleration, and s is the distance traveled. The block starts from rest (u=0), the acceleration is 'a' from Step 2, and the distance is 2.00 m. Substitute all known values to find the final speed v at the bottom of the incline.

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Key Concepts

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

Understanding Newton's Second Law in Physics
Newton's second law of motion is foundational when analyzing the movement of objects. In essence, it states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass. This relationship is succinctly expressed in the equation: \[\begin{equation}F = ma,\end{equation}\]where 'F' represents the net force, 'm' is the object's mass, and 'a' is the acceleration. When investigating motion on an incline, this law becomes particularly significant, as it allows us to calculate an object's acceleration only with knowledge of the forces acting along the incline.A core aspect of applying Newton's second law is breaking down forces into their components. For example, when dealing with an incline, the force of gravity can be resolved into two components: parallel and perpendicular to the plane. This simplification is crucial because it enables us to focus on the forces that directly influence the object's acceleration down the incline without getting bogged down by orthogonal forces which don't affect it directly.
Applying Kinematic Equations in Motion Analysis
Kinematic equations serve as a powerful set of tools for describing the motion of objects. They relate the variables of motion—distance (s), initial velocity (u), final velocity (v), acceleration (a), and time (t)—without considering the forces that cause this motion.One such equation is: \[\begin{equation}v^2 = u^2 + 2as, \end{equation}\]which is instrumental when you need to find the final speed of an object, like a block sliding down an incline, without accounting for the effect of time. Practically, this equation allows students to directly calculate the speed of the block at the bottom of the incline using its acceleration and the distance traveled, which is especially helpful in problems where time is an unknown variable or is not required to find the solution.
Incline Plane Physics and Motion
In physics, an inclined plane simplifies the study of forces and motion along a slope. The physics at play on an incline is quintessential when analyzing how objects slide or roll downward under the influence of gravity.Understanding motion on an incline requires acknowledgement of how forces are affected by the angle of the slope, \[\begin{equation}\theta, \end{equation}\]and how this angle alters the gravitational force experienced by the object. When no friction is present, as in our exercise, the component of gravity causing the block to accelerate is solely the one acting parallel to the incline's surface. This force can be calculated with the expression: \[\begin{equation}mg\sin(\theta).\end{equation}\]Where 'm' stands for mass, 'g' for standard gravitational acceleration, and \[\begin{equation}\sin(\theta)\end{equation}\]is the sine of the inclination angle. Thus, the steeper the incline, the stronger this component will be, leading to greater acceleration of the object.Communicating these concepts with clarity involves emphasizing how each variable interacts and influences object behavior on an incline, enabling students to predict and calculate motion using fundamental physics principles.

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Most popular questions from this chapter

To prevent a box from sliding down an inclined plane, student A pushes on the box in the direction parallel to the incline, just hard enough to hold the box stationary. In an identical situation student B pushes on the box horizontally. Regard as known the mass \(m\) of the box, the coefficient of static friction \(\mu_{s}\) between box and incline, and the inclination angle \(\theta .\) (a) Determine the force A has to exert. (b) Determine the force \(B\) has to exert. (c) If \(m=2.00 \mathrm{kg}, \theta=25.0^{\circ},\) and \(\mu_{s}=0.160,\) who has the easier job? (d) What if \(\mu_{s}=0.380 ?\) Whose job is easier?

An electron of mass \(9.11 \times 10^{-31} \mathrm{kg}\) has an initial speed of \(3.00 \times 10^{5} \mathrm{m} / \mathrm{s}\) . It travels in a straight line, and its speed increases to \(7.00 \times 10^{5} \mathrm{m} / \mathrm{s}\) in a distance of \(5.00 \mathrm{cm} .\) Assuming its acceleration is constant, (a) determine the force exerted on the electron and (b) compare this force with the weight of the electron, which we neglected.

An 8.40 -kg object slides down a fixed, frictionless inclined plane. Use a computer to determine and tabulate the normal force exerted on the object and its acceleration for a series of incline angles (measured from the horizontal) ranging from \(0^{\circ}\) to \(90^{\circ}\) in \(5^{\circ}\) increments. Plot a graph of the normal force and the acceleration as functions of the incline angle. In the limiting cases of \(0^{\circ}\) and \(90^{\circ},\) are your results consistent with the known behavior?

If a man weighs 900 \(\mathrm{N}\) on the Earth, what would he weigh on Jupiter, where the acceleration due to gravity is 25.9 \(\mathrm{m} / \mathrm{s}^{2}\) ?

A woman at an airport is towing her 20.0 -kg suitcase at constant speed by pulling on a strap at an angle \(\theta\) above the horizontal (Fig. P5.40). She pulls on the strap with a \(35.0-\mathrm{N}\) force, and the friction force on the suitcase is 20.0 \(\mathrm{N}\) . Draw a free-body diagram of the suitcase. (a) What angle does the strap make with the horizontal? (b) What normal force does the ground exert on the suitcase?

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