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Chapter 5: Problem 8

Which of the following statements are incorrect? (a) If there were no friction, work needed to be done to move a body up an inclined plane will be zero. (b) If there were no friction, the moving vehicles could not be stopped even by applying hard brakes. (c) As the angle of inclination is increased, the normal reaction on the body placed on it increases. (d) A duster weighing \(0.5 \mathrm{~N}\) is pressed against a vertical board with a force of \(11 \mathrm{~N}\). If the coefficient of friction is \(0.5\), the work donc in rubbing it upwards through a distanec of \(10 \mathrm{~cm}\) is \(0.55 \mathrm{~J}\).

Short Answer

Expert verified
Statements (a) and (c) are incorrect.

Step by step solution

01

Analyze Statement (a)

For statement (a), consider the concept of work needed to move a body on an inclined plane without friction. Without friction, the only force opposing the motion is due to gravity. Thus, work must be done against the component of gravitational force acting down the inclined plane. Therefore, this statement is incorrect because work would be required.
02

Analyze Statement (b)

For statement (b), examine the role of friction in stopping vehicles. Without friction, there would be no force to counteract the motion when brakes are applied, making it impossible to stop the vehicle. Hence, this statement is correct because stopping a vehicle requires friction.
03

Analyze Statement (c)

For statement (c), consider the normal reaction on an inclined plane. As the angle of inclination increases, the component of gravitational force perpendicular to the plane decreases, leading to a reduction in the normal force. Therefore, this statement is incorrect as increasing the angle decreases the normal reaction.
04

Analyze Statement (d)

For statement (d), calculate the work done using the formula for work done against friction: \( W = f \cdot d \), where \( f = \mu \cdot N \). Here, \( N = 11 \text{ N} \), \( \mu = 0.5 \), and \( d = 0.1 \text{ m} \). The frictional force \( f = 0.5 \times 11 = 5.5 \text{ N} \). Therefore, \( W = 5.5 \times 0.1 = 0.55 \text{ J} \), which matches the statement. Hence, this statement is correct.

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Key Concepts

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

Inclined Plane Mechanics
Understanding inclined plane mechanics is essential to grasp how objects move on surfaces that are not horizontal. An inclined plane is simply a flat surface raised at an angle relative to the ground. Gravity acts on any object placed on this inclined surface, and the gravitational force can be broken down into two components:
  • One that acts parallel to the plane, pulling the object downwards.
  • Another that acts perpendicular to the plane, pushing the object against the surface.
When an object moves up an inclined plane, work must be done against the gravitational force component acting downward along the plane. Without the presence of friction, this work is solely against gravity, making option (a) incorrect because moving an object still requires work, albeit less than if there was friction. Inclined planes are widely used to reduce the effort needed to raise objects, as they spread the necessary workload over a longer distance but with less force. This is why ramps are common in loading docks and wheelchair-accessible structures.
Work Done Against Friction
When moving an object across a surface, work done against friction is an important factor to consider. This kind of work occurs because friction opposes the motion of the object, requiring additional force to overcome it. The formula for calculating work done against friction is: \( W = f \cdot d \), where \( f \) is the frictional force and \( d \) is the distance over which the force is applied. The frictional force \( f \) is calculated using the normal force and the coefficient of friction \( \mu \) by the equation: \( f = \mu \cdot N \). In our original exercise, a force of 11 N presses a duster against a board, with a coefficient of friction of 0.5. Calculating \( f \), we get 5.5 N. If the duster moves 10 cm (or 0.1 meters), the work done against friction is indeed 0.55 J, confirming the statement's correctness. Such calculations are crucial in designing systems like brake mechanisms in vehicles. It is the friction that provides the stopping force when brakes are applied, explaining why option (b) is correct. Without friction, even hard-applied brakes won't stop a vehicle.
Normal Force in Physics
Normal force in physics refers to the force exerted by a surface perpendicular to an object resting on it. It is crucial for understanding interactions between surfaces, particularly when calculating friction, as it directly influences the frictional force. On an inclined plane, normal force changes with the angle of inclination. As the angle increases, the component of gravitational force acting perpendicular to the plane decreases, which causes a reduction in the normal force. Consequently, this can affect the friction experienced by the object, since friction is dependent on normal force. In the exercise, the statement regarding the increase of normal reaction with angle is incorrect; in fact, normal force decreases as inclination angle increases. Normal force is not just essential in dealing with inclined planes; it also plays a significant role in everyday situations. For example, when you sit on a chair, the chair exerts an upward normal force equal to your weight, keeping you in balance. These principles help explain various physical scenarios and are foundational to understanding how objects interact with surfaces.

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Most popular questions from this chapter

In certain position \(X\), kinetic energy of a particle is \(25 \mathrm{~J}\) and potential energy is \(-10 \mathrm{~J} .\) In another position \(Y\), kinetic energy of the particle is \(95 \mathrm{~J}\) and the potential energy is \(-25 \mathrm{~J}\). During the displacement of the particle from \(X\) to \(Y\) (a) Net work done by all the forces is \(-70 \mathrm{~J}\). (b) Work done by the conservative forces is \(-15 \mathrm{~J}\). (c) Work done by all the forces besides the conservative forces is \(55 \mathrm{~J}\). (d) Work done by the conservative forces equals work done by the non- conservative forces.

Statement-1: The work done by all forces on a system equals to the change in kinetic energy of that system. This statement is true even if nonconservative forces act on the system. Statement-2: The total work done by internal forces may be positive.

Statement-1: \(\frac{d U}{d x}<0\) implies that corresponding conservative force is directed along positive direction of \(x\) -axis. Statement-2: Conscrvative force is dirccted from region of high polential cnergy to region of low potential energ \(y\).

\(\Lambda\) chain of mass \(m\) and length \(l\) is held vertical, such that its lower end just touches the floor. I released from rest. Find the force exeried by the chain on the table when upper end is about to hit the foor. Solution Force \(F\) exerted by chain consists of two components (a) \(F_{1}\) weight of the fallen portion of the chain, (b) \(F_{2}\) thrust of the falling part of chain. Now consider an clement of chain of length \(d y\) at a height \(y\) from the floor. It will strike the floor with a velocity \(v-\sqrt{2 g y}\). Thus we have, \(\Gamma_{1}=\lambda y g\) Here \(\lambda\) is the mass per unit length of chain and \(\Gamma_{2}-v_{\mathrm{rel}} \frac{d m}{d t}\) We have \(v_{\mathrm{rel}}=v \quad\) and \(\quad d m=\lambda d x \quad \therefore F=-v \frac{\lambda d x}{d t}-\lambda v^{2}\) 'Ihe force exerted by chain on the floor,$$ F=F_{1}+F_{2}=\lambda y g+\lambda v^{2}-\lambda y g+\lambda(\sqrt{2 g y})^{2}=\lambda y g+2 \lambda g=3 \lambda y g $$ When upper end is about to hit the floor, \(y=l\) \(\therefore \quad F=3 \lambda / g=3 m g\)

In casc of a simple pendulum of length \(L:\) (a) The maximum possible velocity \((v)\) at lowcst point for oscillations is \(\sqrt{2 g L}\). (b) The pendulum may leave the circle after reaching a certain height if \(v\) lies between \(\sqrt{2 g L}\) and \(\sqrt{5 g L}\) (c) The pendulum will loop if \(v>\sqrt{5 g L}\). (d) None
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