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Chapter 8: Problem 30

A particle is released one by one from the top of two inclined rough surfaces of height \(h\) each. The angles of inclination of the two planes are \(30^{\circ}\) and \(60^{\circ}\), respectively. All other factors (e.g., coefficient of friction, mass of block, etc.) are same in both the cases. Let \(K_{1}\) and \(K_{2}\) be the kinetic energies of the particle at the bottom of the plane in the two cases. Then (1) \(K_{1}=K_{2}\) (2) \(K_{1}>K_{2}\) (3) \(K_{1}

Short Answer

Expert verified
Kinetic energy comparison yields \(K_1 < K_2\).

Step by step solution

01

Analyze Energy Conservation

We begin by applying the principle of conservation of energy. Initially, the particles have potential energy given by \(PE = mgh\), where \(m\) is mass, \(g\) is acceleration due to gravity, and \(h\) is the height. This potential energy is converted into kinetic energy (\(KE\)) and work done against friction. The equation is \(mgh = KE + W_{f}\).
02

Identify Work Done Against Friction

The work done against friction \(W_f\) depends on the path length \(L\) of the plane, which relates to \(h\) as \(L = \frac{h}{\sin \theta}\), where \(\theta\) is the angle of inclination. The frictional force \(f = \mu mg \cos \theta\) acts over this length. Hence, \(W_f = \mu mg \cos \theta \cdot \frac{h}{\sin \theta}\).
03

Calculation of Kinetic Energy for Each Incline

We substitute \(W_f\) into the energy equation. Solving for \(KE\), we get:\[KE = mgh - \mu mg \frac{h \cos \theta}{\sin \theta} = mg h \left( 1 - \mu \cot \theta \right)\]For \(\theta_1 = 30^\circ\) and \(\theta_2 = 60^\circ\), calculate \(KE_1\) and \(KE_2\) using \(\cot 30^\circ = \sqrt{3}\) and \(\cot 60^\circ = \frac{1}{\sqrt{3}}\).
04

Compare the Kinetic Energies

For the given angles, \(KE_1 = mgh(1 - \mu \sqrt{3})\) and \(KE_2 = mgh(1 - \mu/\sqrt{3})\). Since \(\sqrt{3} > 1/\sqrt{3}\), it follows that:\[KE_1 < KE_2\]Thus, the kinetic energy at the bottom of the plane with a \(30^\circ\) incline is less than the kinetic energy at the bottom of the plane with a \(60^\circ\) incline.

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Key Concepts

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

Inclined Planes
Inclined planes are simple yet fundamental structures used to study motion. Imagine a flat surface tilted at an angle, which allows objects to slide down due to gravity.

When dealing with physics problems, identifying the angle of inclination is crucial. This angle, denoted as \( \theta \), directly affects how gravity splits into two components:
  • One component parallel to the plane, causing the downward slide.
  • Another perpendicular, affecting the normal force.

In our exercise, we observe two inclined planes with angles of 30掳 and 60掳. The steeper the incline (greater \( \theta \)), the greater the component of gravity aiding the slide. This results in varied speeds at different angles, making inclined planes an ideal subject for learning about forces and energy conservation.
Kinetic Energy
Kinetic energy represents the energy of motion. It quantifies how much work an object can perform due to its velocity, calculated as \( KE = \frac{1}{2}mv^2 \), where \( m \) is mass and \( v \) is velocity.

In our analysis, the potential energy from height \( h \) is converted into kinetic energy. Initially, the particle has potential energy \( PE = mgh \), which at the bottom becomes kinetic energy subtracted by work done against friction.

The steeper the plane (larger \( \theta \)), the lesser the friction's effect via a shorter path. More potential energy is then converted to kinetic energy, explaining why \( KE_1 < KE_2 \). This illustrates the principle of energy conservation in mechanics, crucial for understanding motion dynamics.
Frictional Force
Friction opposes motion, and on inclined planes, it acts parallel to the surface, resisting the slide. This force depends on the normal component of gravity: the greater the angle, the smaller the normal force.

Frictional force \( f \) is determined by \( f = \mu mg \cos \theta \), where \( \mu \) is the coefficient of friction and \( \theta \) the angle of inclination.

In our exercise, we observe the frictional work performed, calculated as \( W_f = \mu mg \frac{h \cos \theta}{\sin \theta} \). As \( \theta \) increases:
  • \( \cos \theta \) decreases, lessening the friction force.
  • \( \sin \theta \) increases, making the path shorter.
Thus, a steeper path allows more conversion of stored energy into kinetic energy, resulting in \( KE_1 < KE_2 \), highlighting how friction impacts energy conservation.

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Most popular questions from this chapter

A system consists of two identical cubes, each of mass \(3 \mathrm{~kg}\), linked together by a compressed weightless spring of force constant \(1000 \mathrm{~N} \mathrm{~m}^{-1}\). The cubes are also connected by a thread which is burnt at a certain moment. At what minimum value of initial compression \(x_{0}\) (in \(\mathrm{cm}\) ) of the spring will the lower cube bounce up after the thread is burnt through?

A chain of length \(l\) and mass \(m\) lies on the surface of a smooth sphere of radius \(R>l\) with one end tied to the top of the sphere. $$ \begin{array}{|c|c|} \hline {\begin{array}{c} \text { Column I } \\ \end{array}} & {\begin{array}{c} \text { Column II } \\ \end{array}} \\ \hline \text { i. } \text {Gravitational potential energy w.r.t. centre of the sphere} & \text { a. } \text{\(\frac{\operatorname{Rg}}{l}\left[1-\cos \left(\frac{l}{R}\right)\right]\)} \\ \hline \text { ii. } \text{The chain is released and slides down, its \(\mathrm{KE}\) when it has slid by \(\theta\)} & \text { b. } \text{\(\frac{2 R g}{l}\left[\sin \left(\frac{l}{R}\right)+\sin \theta-\sin \left(\theta+\frac{l}{R}\right)\right]\)} \\ \hline \text { iii. } \text{The initial tangential acceleration} & \text { c. } \text{\(\frac{M R^{2} g}{l} \sin \left(\frac{l}{R}\right)\)}\\\ \hline \text { iv. } \text{The radial acceleration \(a_{r}\)} & \text { d. } \text{\(\frac{M R^{2} g}{l}\left[\sin \left(\frac{l}{R}\right)+\sin \theta-\sin \left(\theta+\frac{l}{R}\right)\right]\)} \\ \hline \end{array} $$

An object of mass \(m\) slides down a hill of arbitrary shape and after travelling a certain horizontal path stops because of friction. The total vertical height descended is \(h\). The friction coefficient is different for different segments for the entire path but is independent of the velocity and direction of motion. The work that a tangential force must perform to return the object to its initial position along the same path is (1) \(\mathrm{mgh}\) (2) \(-m g h\) (3) \(-2 m g h\) (4) \(2 \mathrm{mgh}\)

A projectile is launched at angle \(\theta\) to the horizontal from point \(L\) and it hits the target \(T\) on level ground. During the entire motion: $$ \begin{array}{|c|c|} \hline {\begin{array}{c} \text { Column I } \\ \end{array}} & {\begin{array}{c} \text { Column II } \\ \end{array}} \\ \hline \text { i. } \text {Magnitude of radial acceleration} & \text { a. } \text{remains constant} \\ \hline \text { ii. } \text{Magnitude of tangential acceleration} & \text { b. } \text{either always decreases or always increases} \\ \hline \text { iii. } \text{Power delivered by gravity} & \text { c. } \text{First increases, then decreases}\\\ \hline \text { iv. } \text{Total acceleration} & \text { d. } \text{First decreases, then increases} \\ \hline \end{array} $$ Now match the given columns and select the correct option from the codes given below. Codes: $$ \begin{array}{lllll} & \text { i. } & \text { ii. } & \text { iii. } & \text { iv. } \\ \text { (1) } &\text { a }& \text { b} & \text { c} & \text { d}\\\ \text { (2) } &\text { c }& \text { d } & \text { c } & \text { a}\\\ \text { (3) } &\text { d }& \text { a } & \text { b } & \text { c }\\\ \text { (4) } &\text { a }& \text { c } & \text { b } & \text { d } \end{array} $$

A car drives along a straight level frictionless road by an engine delivering constant power. Then velocity is directly proportional to \(\begin{array}{ll}\text { (1) } t & \text { (2) } \frac{1}{\sqrt{t}}\end{array}\) (3) \(\sqrt{t}\) (4) None of these
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