/*! 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 53 A bicyclist comes to a skidding ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 53

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 \((\mathrm{A})+2000 \mathrm{~J}\) (B) \(-200 \mathrm{~J}\) (C) Zero (D) \(-20,000 \mathrm{~J}\)

Short Answer

Expert verified
The work done by the bicycle on the road is \(W = -2000 \mathrm{~J}\), which corresponds to option (A) +2000 J.

Step by step solution

01

Recall the work-energy theorem

Recall the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy: \[W = \Delta KE\] where \(W\) is work, and \(\Delta KE\) is the change in kinetic energy of the object.
02

Use the formula for work

The formula to calculate work is: \[W = F \cdot d \cdot cos(\theta)\] where \(W\) is work, \(F\) is the force, \(d\) is the distance, and \(\theta\) is the angle between the force and displacement. In this case, the force is opposite to the direction of motion (directly opposed to the motion). Therefore, the angle \(\theta\) between force and motion is \(\mathrm{180^{\circ}}\) or \(\pi\) radians. So, the cosine of \(\theta\) is \(\mathrm{cos(180^{\circ})} = \mathrm{-1}\).
03

Calculate the work done by the bicycle on the road

Now, plug the values into the work formula: \[W = F \cdot d \cdot cos(\theta) = 200 \mathrm{~N} \cdot 10 \mathrm{~m} \cdot (-1) = -2000 \mathrm{~ J}\] We find that the work done by the bicycle on the road is \(W = -2000 \mathrm{~J}\). Thus, the correct answer is (A) +2000 J.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two identical beads of \(m=100 \mathrm{~g}\) are connected by an inextensible massless string that can slide along the two arms \(A C\) and \(B C\) of a rigid smooth wireframe in a vertical plane. If the system is released from rest, the kinetic energy of the first particle when they have moved by a distance of \(0.1 \mathrm{~m}\) is \(16 x \times 10^{-3} \mathrm{~J}\). Find the value of \(x .\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)

The potential energy of a particle of mass \(m\) is given by \(U=\frac{1}{2} k x^{2}\) for \(x<0\) and \(U=0\) for \(x \geq 0 .\) If total mechanical energy of the particle is \(E\). Then its speed at \(x=\sqrt{\frac{2 E}{k}}\) is (A) Zero (B) \(\sqrt{\frac{2 E}{m}}\) (C) \(\sqrt{\frac{E}{m}}\) (D) \(\sqrt{\frac{E}{2 m}}\)

Two identical balls are projected, one vertically up and the other at an angle of \(30^{\circ}\) with the horizontal, with same initial speed. The potential energy at the highest point is in the ratio (A) \(4: 3\) (B) \(3: 4\) (C) \(4: 1\) (D) \(1: 4\)

When a body moves in a circle, the work done by the centripetal force is always (A) \(>0\) (B) \(<0\) (C) Zero (D) None of these

Figure \(4.22\) shows a massless spring fixed at the bottom end of an inclined of inclination \(37^{\circ}\left(\tan 37^{\circ}=\right.\) \(3 / 4\) ). A small block of mass \(2 \mathrm{~kg}\) start slipping down the incline from a point \(4.8 \mathrm{~m}\) away from free end of spring. The block compresses the spring by \(20 \mathrm{~cm}\), stops momentarily and then rebounds through a distance \(1 \mathrm{~m}\) up the inclined, then \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\). (A) Coefficient of friction between block and inclined is \(0.5\). (B) Coefficient of friction between block and inclined is \(0.75\). (C) Value of spring constant is \(1000 \mathrm{~N} / \mathrm{m}\). 69 (D) Value of spring constant is \(2000 \mathrm{~N} / \mathrm{m}\).
See all solutions

Recommended explanations on Physics Textbooks

Waves Physics

Read Explanation

Particle Model of Matter

Read Explanation

Radiation

Read Explanation

Geometrical and Physical Optics

Read Explanation

Rotational Dynamics

Read Explanation

Space Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.