/*! 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}

91Ó°ÊÓ

The string in Fig. 8-38 is L=120cmlong, has a ball attached to one end, and is fixed at its other end. The distancedfrom the fixed end to a fixed peg at point P is 75.0cm. When the initially stationary ball is released with the string horizontal as shown, it will swing along the dashed arc. What is its speed when it reaches (a) its lowest point and (b) its highest point after the string catches on the peg?

Short Answer

Expert verified
  1. The speed when it reaches its lowest point 4.85m/s.
  2. The speed when it reaches its highest point is 2.42m/s.

Step by step solution

01

Given data:

Length of the string, L=120cm=1.20m

The distance from the fixed end to a fixed peg at point P, d=75.0cm=0.75m

02

Understanding the concept:

The law of conservation of energy states that energy cannot be created or destroyed - only converted from one form of energy to another. This means that the system always has the same amount of energy unless it is added from outside.

03

(a) Calculate the speed when it reaches its lowest point:

As the string reaches its lowest point, its original potential energy is,

U=mgL

(Measured relative to the lowest point) is converted into kinetic energy. Thus,

mgL=12mv2

v=2gL(1)

Here,

The acceleration due to gravity, g=9.80m/s2

Substitute known values in the above equation.

v=2(9.80m/s2)(1.20m)=4.85m/s

Hence, the speed when it reaches its lowest point 4.85m/s.

04

(b) Calculate the speed when it reaches its highest point:

In this case, the total mechanical energy is shared between kinetic 12mvb2and potential mgyb. Note that yb=2r

Where the distance,

r=L−d=(1.20−0.75)m=0.450m

Energy conservation leads to:

mgL=12mvb2+mgyb12mvb2=mgL−mgybvb2=2mg(L−yb)m

vb=2g(L−2r)=2(9.80m/s2)(1.20m−2(0.450m))=(19.6m/s2)(0.3m)=5.88m2/s2

vb=2.42m/s

Here, the speed when it reaches its highest point is 2.42m/s.

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

A 3.2 kgsloth hangs 3.0 mabove the ground. (a) What is the gravitational potential energy of the sloth-Earth system if we take the reference point y=0to be at the ground? If the sloth drops to the ground and air drag on it is assumed to be negligible, what are the (b) kinetic energy and (c) speed of the sloth just before it reaches the ground?

Figure shows a plot of potential energy Uversus position xof ag particle that can travel only along an xaxis. (Nonconservative forces are not involved.) Three values are, UA= 15.0J, UB = 35.0 Jand UC = 45.0 J. The particle is released at x= 4.5 mwith an initial speed of, headed in the negative xdirection. (a) If the particle can reach x = 1.0 m, what is its speed there, and if it cannot, what is its turning point? What are the (b) magnitude and (c) direction of the force on the particle as it begins to move to the left of x = 4.0 m? Suppose, instead, the particle is headed in the positive xdirection when it is released at x = 4.5 mat speed 7.0 m/s. (d) If the particle can reach x= 7.0 m, what is its speed there, and if it cannot, what is its turning point? What are the (e) magnitude and (f) direction of the force on the particle as it begins to move to the right of?

A stone with a weight of 52.9 Nis launched vertically from ground level with an initial speed of 20.0 m/s, and the air drag on it is 0.265 Nthroughout the flight. What are (a) the maximum height reached by the stone and (b) Its speed just before it hits the ground?

In Fig. 8-25, a block slides along a track that descends through distance h. The track is frictionless except for the lower section. There the block slides to a stop in a certain distance Dbecause of friction. (a) If we decrease h, will the block now slide to a stop in a distance that is greater than, less than, or equal to D? (b) If, instead, we increase the mass of the block, will the stopping distance now be greater than, less than, or equal to D?

From the edge of a cliff, a 0.55 kgprojectile is launched with an initial kinetic energy of 1550 J. The projectile’s maximum upward displacement from the launch point is +140 m. What are the (a) horizontal and (b) vertical components of its launch velocity? (c) At the instant the vertical component of its velocity is 65 m/s, what is its vertical displacement from the launch point?

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.