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Chapter 2: Problem 13

\(\Lambda\) body slides down a smooth inclined plane covering equal distance \(s\) in consecutive intervals of time \(t_{1}\) and \(t_{2} .\) Find the inclination of the plane.

Short Answer

Expert verified
The approach should be rechecked as error implies constraints induce non-linear angle detection under constant separation.

Step by step solution

01

Understand the Problem

We are given that a body slides down an inclined plane, covering equal distances in consecutive time intervals. We need to find the inclination of the plane in terms of these given values.
02

Define Variables and Equations

Let the initial velocity of the body be zero and the inclination angle be \( \theta \). The acceleration \( a \) along the incline is caused by gravity: \( a = g \sin \theta \). We also know the distances are equal: \( s = \frac{1}{2}at_{1}^2 = \frac{1}{2}a(t_{1} + t_{2})^2 - \frac{1}{2}at_{1}^2 \).
03

Express Equal Distances Mathematically

Set the distances equal from the above equations: \( \frac{1}{2}a t_{1}^2 = \frac{1}{2}a[t_{1} + t_{2}]^2 - \frac{1}{2}a t_{1}^2 \). Simplify this to find: \( at_{1}^2 = a(t_{1}^2 + 2t_{1}t_{2} + t_{2}^2 - t_{1}^2) \), leading to \( at_{1}^2 = 2a t_{1} t_{2} + a t_{2}^2 \).
04

Solve for Acceleration in Terms of Given Values

Reorganize the terms in the equation: \( a t_{1}^2 = 2at_{1}t_{2} + at_{2}^2 \). Simplify to \( at_{1}^2 - 2at_{1}t_{2} = at_{2}^2 \). Further reducing gives \( a(t_{1} - t_{2})^2 = 0 \), which implies \( t_{1} = t_{2} \).
05

Substitute Solutions to Find Angle Inclination

The result \( t_{1} = t_{2} \) seems inconsistent with non-equal time intervals required for constant acceleration constraints. Recheck assumptions leading to incorrect steps or inconsistent checking until finding: Given \( a = g \sin \theta \), establish altitude/angle by solving algebraic concatenation consistently.

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Key Concepts

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

kinematics
Kinematics is the branch of physics that deals with the motion of objects, without considering the forces that cause this motion. In the context of an inclined plane, it involves understanding how objects move along the plane under the influence of gravity.

Some key concepts include:
  • Distance and Displacement: Distance is a scalar quantity that represents how much ground an object has covered during its motion. Displacement, on the other hand, is a vector quantity that points from the initial to the final position of the object.
  • Speed and Velocity: Speed is the rate of change of distance with time, while velocity is speed in a given direction. Like displacement, velocity is a vector quantity.
  • Acceleration: This is the rate of change of velocity with time. On an inclined plane, acceleration can vary depending on the angle of the plane and the forces acting on the object.
In the problem of an object sliding down an inclined plane, we're focused on uniform acceleration caused by gravity. This is modeled using the equations of motion in kinematics:

\[ s = ut + \frac{1}{2} a t^2 \] where \(s\) is the displacement, \(u\) is the initial velocity, \(a\) is acceleration, and \(t\) is time.
acceleration on inclined planes
When dealing with inclined planes, acceleration plays a crucial role in determining how objects move. The acceleration of a body on an inclined plane is influenced by gravity and the angle of the plane.

Gravity acts downwards, and when on an incline, only a component of this force causes acceleration down the plane.

  • Acceleration Due to Gravity: The force of gravity is split into two components - one acting perpendicular to the plane and one parallel. It is the parallel component that influences the acceleration of an object down the slope.
  • Formula: The parallel component of gravity causes the acceleration \( a = g \sin \theta \), where \( g \) is the acceleration due to gravity and \( \theta \) is the angle of the incline.
In the provided exercise, acceleration is crucial to determining how an object covers equal distances over consecutive time intervals. Understanding that \( a \) is derived from these components helps in accurately predicting the object's motion.
trigonometric functions in physics
Trigonometric functions like sine, cosine, and tangent are essential tools in physics to solve problems involving angles, particularly with inclined planes.

They form the basis for decomposing vectors or forces that act at an angle, making them especially important in inclined plane scenarios.

  • Sine Function: Sine is often used to calculate the component of gravitational force acting parallel to the inclined plane. For example, \( \sin \theta \) is used in calculating \( a = g \sin \theta \).
  • Cosine Function: This function helps determine the component of gravity acting perpendicular to the plane, which is essential for understanding normal force.
  • Angle of Inclination: In solving these problems, knowing the angle \( \theta \) is crucial. It allows us to correctly compute the acceleration, helping us predict motion accurately.
The exercise emphasizes the use of trigonometry to relate the angle of the plane with the motion of the object, showcasing the integral role these functions play in solving physics problems effectively.

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Most popular questions from this chapter

\(\Lambda\) wooden block of mass \(10 \mathrm{~g}\) is dropped from the top of a cliff \(100 \mathrm{~m}\) high. Simultaneously, a bullet of mass \(10 \mathrm{~g}\) is fired from the foot of the cliff upwards with a velocity of \(100 \mathrm{~m} / \mathrm{s}\). Find (a) where and after what time will they meet. (b) if the bullet after striking the block, gets embedded in it, how high will it rise above the cliff before it starts falling? ( Take \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\) )

If velocity of a particle moving along a straight line changes with time as \(v(\mathrm{~m} / \mathrm{s})-4 \sin \frac{\pi}{2} t\), its average velocity over time interval \(t=0\) to \(t=(2 n-1) 2 \mathrm{sec},,(n\) being any \(+\) ve integer \()\) is (a) \(\frac{16}{n(2 n 1)} \mathrm{m} / \mathrm{s}\) (b) \(\frac{8}{\pi(2 n-1)} \mathrm{m} / \mathrm{s}\) (c) zero (d) \(\frac{16(2 n \mathrm{l})}{\pi} \mathrm{m} / \mathrm{s}\)

Mark the correct statements for a particle going on a strainght line: (a) If the velocity and acceleration have opposite sign, the object is slowing down. (b) If the position and velocity have opposite sign, the particle is moving towards the origin. (c) If the velocity is zero at an instant, the acceleration should also be zero at that instant. (d) If the velocity is zero for a time interval, the acceleration is zero at any instant within the time interval.

\(\Lambda\) particle is projected vertically upwards with velocity \(40 \mathrm{~m} / \mathrm{s}\). Find the displacement and distance travelled by the particle in (a) \(2 \mathrm{~s}\), (b) \(4 \mathrm{~s}\) and (c) 6 s (lake \(\left.\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) Solution Herc \(u\) is positive (upwards) and \(a\) is negative (downwards). So, first we will find \(t_{o}\), the time when velocity becomes rero. \(t_{\mathrm{s} 1}-\left|\frac{u}{\alpha}\right|-\frac{40}{10}-4 s\) (a) \(tt_{\mathrm{a}}\) as \(t=6 \mathrm{~s} .\) Hence \(d>s\) \(s-40 \times 6-\frac{1}{2} \times 10 \times 36-60 \mathrm{~m}\) whilc \(\quad d-\left|\frac{u^{2}}{2 a}\right|+\frac{1}{2}\left|a\left(t-t_{v}\right)^{2}\right|-\frac{(40)^{2}}{2 \times 10}+\frac{1}{2} \times 10 \times(6-4)^{2}-100 \mathrm{~m}\).

A particle moves in a straight line with initial velocity \(4 \mathrm{~m} / \mathrm{s}\) and a constant acecleration of \(6 \mathrm{~m} / \mathrm{s}^{2}\). Find the average velocity of the particle in a time interval from (a) \(t=0\) to \(t=2 s\) (b) \(t=2 s\) to \(t=4 s\).
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