/*! 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} Q12 P A man whose mass is 80 kg and a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A man whose mass is 80kg and a woman whose mass is 50kgsit at opposite ends of a canoe 5m long, whose mass is 30kg. (a) Relative to the man, where is the mass of the system consisting of man-woman, and canoe? (Hint: Choose a specific coordinate system with a specific origin.) (b) Suppose that the man moves quickly to the center of the canoe and sits down there. How far does the canoe move in the water? Explain your work and your assumptions.

Short Answer

Expert verified
  1. The center of mass of the system consisting of man-woman, and canoe, Relative to the man is 2.03m
  2. The canoe moves 1.72m in the downward direction.

Step by step solution

01

Identification of given data

  • The mass of a man is m1=80kg
  • The mass of a woman is m2=50kg
  • The mass of a canoe is m3=30kg
  • The length of a canoe is 5m long
02

Concept of the mass of the system

The mass of the system is defined by considering the average positions of all the objects acting in the system.

03

(a) Determination of the mass of the system consisting of man-woman, and canoe, Relative to the man

The center of mass of the system can be,

MxCM=m1x1+m2x2+m3x3

Where,

x1=0m,x2=2.5m,x3=5mM=m1+m2+m3=80+50+30=160kg

Substitute these values in above expression,

MxCM=m1x1+m2x2+m3x3=(80×0)+(30×2.5)+(50×5)(kg·m)=325M·1kg·m1kg

Substituting M values in above expression

xCM=325160·1kg·m1kg=2.03·1m=2.03m

Hence, the center of mass of the system is2.03m

04

(b) Determination of the new center of mass

We know that,

MxCM=m1x1+m2x2+m3x3=(80×2.5)MxCM=200M·1kg·m1kg

Substituting M values in above expression

xCM=2001601kg·m1kgxCM=1.25m

The distanceto find how the canoe moved will be foundby;

=2.5m+1.25m-2.03m=1.72m

Here the negative sign indicates the downward movement of the canoe. The canoe moves 1.72min the downward direction in the water when man moves quickly to the center of the canoe and sits down there.

Hence, the canoe moves 1.72min the downward direction.

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 91Ó°ÊÓ!

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

Show that moment of inertia of a disk of mass M and radius R is 12MR2. Divide the disk into narrow rings, each of radius r and width dr. The contribution I of by one of these rings is r2dm, where dm is amount of mass contained in that particular ring. The mass of any ring is the total mass times the fraction of the total area occupied by the area of the ring. The area of this ring is approximately 2Ï€°ù»å°ù. Use integral calculus to add up all the calculations.

A runner whose mass is 50kgaccelerates from a stop to a speed of 10m/sin 3s. (A good sprinter can run 100min about 10s, with an average speed of 10m/s.) (a) What is the average horizontal component of the force that the ground exerts on the runner’s shoes? (b) How much displacement is there of the force that acts on the sole of the runner’s shoes, assuming that there is no slipping? Therefore, how much work is done on the extended system (the runner) by the force you calculated in the previous exercise? How much work is done on the point particle system by this force? (c) The kinetic energy of the runner increases—what kind of energy decreases? By how much?

Discuss qualitatively the motion of the atoms in a block of steel that falls onto another steel block. Why and how do large-scale vibrations damp out?

A runner whose mass is 50 kgaccelerates from a stop to a speed of10 m / s in 3 s. (A good sprinter can run100 m in about 10 s, with an average speed of 10 m / s.) (a) What is the average horizontal component of the force that the ground exerts on the runner’s shoes? (b) How much displacement is there of the force that acts on the sole of the runner’s shoes, assuming that there is no slipping? Therefore, how much work is done on the extended system (the runner) by the force you calculated in the previous exercise? How much work is done on the point particle system by this force? (c) The kinetic energy of the runner increases—what kind of energy decreases? By how much?

Question: Under what conditions does the energy equation for the point particle system differ from the energy equation for the extended system? Give two examples of such a situation. Give one example of a situation where the two equations look exactly alike.

See all solutions

Recommended explanations on Physics Textbooks

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.