/*! 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 23 A bicyclist comes to a skidding ... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 23

A bicyclist comes to a skidding stop in \(10 \mathrm{~m}\). During this process, the force on the bicycle due to the road is \(200 \mathrm{~N}\) and is directly opposed to the motion. The work done by the cycle on the road is \(\quad\) [NCERT Exemplar] (a) \(+2000 \mathrm{~J}\) (b) - \(200 \mathrm{~J}\) (c) zero (d) \(-20,000 \mathrm{~J}\)

Short Answer

Expert verified
The work done by the cycle on the road is -2000 J; option (d) is correct.

Step by step solution

01

Understanding Work Done

Work done by a force is defined as the product of the force applied and the displacement in the direction of the force. It can be given by the formula:\[ W = F \times d \times \cos(\theta) \]where \( W \) is work done, \( F \) is force, \( d \) is displacement, and \( \theta \) is the angle between the force and displacement vector.
02

Analyze the Given Information

In this problem, the force \( F = 200 \ \text{N} \) (opposing the motion), and the displacement \( d = 10 \ \text{m} \). The force on the bicycle is directly opposed to the direction of displacement, so the angle \( \theta = 180^\circ \).
03

Calculate Cosine of the Angle

Find \( \cos(180^\circ) \), which is \(-1\). This is because the force is directly opposite to the direction of the displacement.
04

Compute the Work Done

Now, substitute the values into the work formula:\[W = 200 \ \text{N} \times 10 \ \text{m} \times \cos(180^\circ) = 200 \times 10 \times (-1) = -2000 \ \text{J}\]Thus, the work done by the cycle on the road is \(-2000 \ \text{J}\).
05

Choose the Correct Option

From the calculated work done, we identify that the answer corresponds to option (d) \(-2000 \ \text{J}\).

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Key Concepts

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

Newton's Laws of Motion
Newton's Laws of Motion are fundamental principles that describe how objects behave when forces act upon them. There are three laws attributed to Sir Isaac Newton, each explaining a different aspect of motion:
  • First Law (Law of Inertia): An object will remain at rest or in uniform motion in a straight line unless acted upon by an external force. This means that if no forces are acting on an object, it will not change its state of motion.

  • Second Law (Law of Acceleration): The acceleration of an object depends on the mass of the object and the amount of force applied. It is given by the equation \(F = ma\), where \(F\) is the force applied, \(m\) is the mass, and \(a\) is the acceleration.

  • Third Law (Action and Reaction): For every action, there is an equal and opposite reaction. This means that if one body exerts a force on another body, the second body will exert a force of equal magnitude and opposite direction on the first body.

In the context of the bicycle problem, the friction force exerted by the ground on the bicycle is an example of the third law, as the bicycle exerts a force back on the ground.
Friction
Friction is a force that opposes the relative motion or tendency of such motion of two surfaces in contact. It is crucial in everyday physics as it enables walking, driving, and many other actions. Without friction, motion would be uncontrollable.

There are different types of friction:
  • Static Friction: The force that needs to be overcome to start moving an object at rest. It occurs when the surfaces in contact are not sliding past each other.

  • Kinetic Friction: The force opposing the motion of two surfaces sliding past one another.

  • Rolling Friction: The force resisting the motion when a body rolls on a surface.

In our bicycle problem, the force opposing the motion is kinetic friction, which brings the bicycle to a stop. This force acts in the opposite direction to the motion, leading to a negative value for the work done.
Displacement in Physics
Displacement in physics refers to the change in position of an object. It is a vector quantity, meaning it has both magnitude and direction. The difference between distance and displacement is crucial:
  • Distance: A scalar quantity that measures the total path length traveled by an object, regardless of direction.

  • Displacement: Only considers the initial and final positions, connecting these points with a straight line in the shortest path possible. Therefore, it can be positive, negative, or zero.

In the problem given, the cyclist's displacement is 10 meters. It's not just about how far the cyclist traveled, but the direct path from the starting point to where the bike stopped. This direct consideration is important when calculating the work done, especially when forces and displacement directions are involved.
Angle between Force and Displacement
The angle between force and displacement is a critical component in determining work done. This angle affects the calculation via the cosine term in the work formula.When calculating work, the formula \(W = F \cdot d \cdot \cos(\theta)\) requires knowing:
  • Force direction: Determines the line of action of force concerning the path taken (displacement).

  • Displacement direction: The straight-line direction between the start and endpoint of movement.

  • \(\theta\) (Theta): The angle between the direction of the applied force and the direction of displacement.

In the exercise, \(\theta = 180^\circ \) signifies that the force is acting in a direction exactly opposite to the displacement. This results in \(\cos(180^\circ) = -1\), leading to negative work done, as the force opposes the movement of the bicycle. Negative work indicates that energy is taken out of the system, evident when stopping an object in motion.

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Most popular questions from this chapter

A man squatting on the ground gets straight up and stand. The force of reaction of ground on the man during the process is \(\quad\) [NCERT Exemplar] (a) constant and equal to \(m g\) in magnitude (b) constant and greater than \(m g\) in magnitude (c) variable but always greater than \(\mathrm{mg}\) (d) at first greater than \(m g\), and later becomes equal to \(m g\)

The potential energy of a particle of mass \(5 \mathrm{~kg}\) moving in the \(x y\)-plane is given by \(U=(-7 x+24 y) \mathrm{J}, x\) and \(y\) being in metre. If the particle starts from rest from origin, then speed of particle at \(t=2 \mathrm{~s}\) is (a) \(5 \mathrm{~ms}^{-1}\) (b) \(01 \mathrm{~ms}^{-1}\) (c) \(17.5 \mathrm{~ms}^{-1}\) (d) \(10 \mathrm{~ms}^{-1}\)

A stone is dropped from the top of a tall tower. The ratio of the kinetic energy of the stone at the end of three seconds to the increase in the kinetic energy of the stone during the next three seconds is (a) \(1: 1\) (b) \(1: 2\) (c) \(1: 3\) (d) \(1: 9\)

A body is falling freely under the action of gravity alone in vacuum. Which of the following quantities remain constant during the fall? [NCERT Exemplar] (a) Kinetic energy (b) Potential energy (c) Total mechanical energy (d) Total linear energy

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