Physics for Scientists and Engineers with Modern Physics
Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
bartleby

Videos

Textbook Question
Book Icon
Chapter 8, Problem 26AP

Review. As shown in Figure P8.26, a light string that does not stretch changes from horizontal to vertical as it passes over the edge of a table. The string connects m1, a 3.50-kg block originally at rest on the horizontal table at a height h = 1.20 m above the floor, to m2, a hanging 1.90-kg block originally a distance d = 0.900 m above the floor. Neither the surface of the table nor its edge exerts a force of kinetic friction. The blocks start to move from rest. The sliding block m1 is projected horizontally after reaching the edge of the table. The hanging block m2 stops without bouncing when it strikes the floor. Consider the two blocks plus the Earth as the system. (a) Find the speed at which m1 leaves the edge of the table. (b) Find the impact speed of m1 on the floor. (c) What is the shortest length of the string so that it does not go taut while m1 is in flight? (d) Is the energy of the system when it is released from rest equal to the energy of the system just before m1 strikes the ground? (e) Why or why not?

Figure P8.26

Chapter 8, Problem 26AP, Review. As shown in Figure P8.26, a light string that does not stretch changes from horizontal to

(a)

Expert Solution
Check Mark
To determine

To determine: The speed at which block of mass m1 leaves the edge of the table.

Answer to Problem 26AP

Answer: The speed at which block of mass m1 leaves the edge of the table is 2.02m/s .

Explanation of Solution

Given Info: A light string that does not stretch changes from horizontal to vertical its passes over the edge of the table. The string connects of mass 3.5kg block originally at rest on the horizontal table at the height 1.20m above the floor to hanging the mass 1.9kg block originally a distance 0.9m above the floor.

Explanation:

The free body diagram of the body of mass m1 and mass m2 .

Physics for Scientists and Engineers with Modern Physics, Chapter 8, Problem 26AP

Figure I

Formula to calculate acceleration of the block by D’Alembert’s principle for mass m1 is,

T=m1a

  • T is the tension in the string.
  • m1 is the mass of block 1 .
  • a is the acceleration of the blocks.

Formula to calculate acceleration of the block by D’Alembert’s principle for mass m2 is,

Tm2g=m2a                                        (I)

  • m2 is the mass of block 2 .
  • g is the acceleration due to gravity.

Substitute m1a for T in equation (I).

m1am2g=m2aa=m2gm2+m1

Formula to calculate the distance covered by the block m2 is,

d=ut+12at2

  • d is the distance covered by the covered by the block m2 .
  • u is the initial velocity of the block of mass m2 .
  • t is the time taken by block of mass m2 to hit the floor.
  • a is the acceleration of the block of mass m2 .

Substitute m2gm2+m1 for a to find s .

d=ut+12×m2gm2+m1×t2

Substitute 3.5kg for m1 , 0 for u , 1.9kg for m2 , 0.9m for d and 9.81m/s2 for g to find t .

0.9m=0×t+12×1.9kg×9.81m/s21.9kg+3.5kg×t2t=0.722s

Formula to calculate the speed of block of mass m1 is,

v=u+at

  • v is the final velocity of block 1 .

Substitute m2gm2+m1 for a to find v .

v=u+m2gm2+m1t

Substitute 0.722s for t , 0 for u , 3.5kg for m1 , 1.9kg for m2 , and 9.81m/s2 for g to find v .

v=0+1.9kg×9.81m/s21.9kg+3.5kg×0.722s=2.02m/s

Conclusion:

Therefore, the speed at which block of mass m1 leaves the edge of the table is 2.02m/s .

(b)

Expert Solution
Check Mark
To determine

To determine: The impact speed of the block of mass m1 on the floor.

Answer to Problem 26AP

Answer: The impact speed of the block of mass m1 on the floor is 5.25m/s .

Explanation of Solution

Given info: A light string that does not stretch changes from horizontal to vertical its passes over the edge of the table. The string connects of mass 3.5kg block originally at rest on the horizontal table at the height 1.20m above the floor to hanging the mass 1.9kg block originally a distance 0.9m above the floor.

Explanation:

Formula to calculate the impact speed of the block of mass m1 on the floor is,

V=v2+2gh

  • V is the impact speed of the block of mass m1 .
  • v is the speed of the block of mass m1 .
  • h is the distance between block of mass m1 and the floo.

Substitute 2.02m/s for v , 1.2m for h and 9.81m/s2 for g to find V .

V=(2.02m/s)2+2×9.81m/s2×1.2m=5.25m/s

Conclusion:

Therefore, the impact speed of the block of mass m1 on the floor is 5.25m/s .

(c)

Expert Solution
Check Mark
To determine

To determine: The shortest length of the string so that it does not go taut while m1 is in flight

Answer to Problem 26AP

Answer: The shortest length of the string so that it does not go taut while m1 is in flight is 1m .

Explanation of Solution

Explanation:

Given information:

A light string that does not stretch changes from horizontal to vertical its passes over the edge of the table. The string connects of mass 3.5kg block originally at rest on the horizontal table at the height 1.20m above the floor to hanging the mass 1.9kg block originally a distance 0.9m above the floor.

Formula to calculate the shortest distance of the string is,

s=v2hg

  • s is the shortest length of the string so that it does not go taut while m1 is in flight.

Substitute 2.02m/s for v , 1.2m for h and 9.81m/s2 for g to find s .

s=2.02m/s2×1.2m9.81m/s2=1m

Conclusion:

Therefore, the shortest length of the string so that it does not go taut while m1 is in flight is 1m .

(d)

Expert Solution
Check Mark
To determine

To determine: Conservation of energy states that the energy of any isolated system remains constant as the energy can neither be created nor destroyed.

Answer to Problem 26AP

Therefore, the energy of the system when it is released from rest equal to the system just before m1 strikes the ground as there is no frictional force present.

Explanation of Solution

Given info:

Explanation: A light string that does not stretch changes from horizontal to vertical its passes over the edge of the table. The string connects of mass 3.5kg block originally at rest on the horizontal table at the height 1.20m above the floor to hanging the mass 1.9kg block originally a distance 0.9m above the floor.

There is no frictional force present here so, any type of losses will not occur here. Thus, the energy of the system when it is released from rest equal to the system just before m1 strikes the ground.

Conclusion:

Therefore, the energy of the system when it is released from rest equal to the system just before m1 strikes the ground as there is no frictional force present.

(e)

Expert Solution
Check Mark
To determine

To determine: The energy of the system when it is released from rest equal to the system just before m1 why or why not.

Answer to Problem 26AP

Therefore, the energy is conserved in the absence of frictional force so, energy of the system when it is released from rest equal to the system just before m1 strikes the ground.

Explanation of Solution

Given info:

Explanation: Conservation of energy is state that the energy of any isolated system is remains constant. So the energy can neither be created nor destroyed.

A light string that does not stretch changes from horizontal to vertical its passes over the edge of the table. The string connects of mass 3.5kg block originally at rest on the horizontal table at the height 1.20m above the floor to hanging the mass 1.9kg block originally a distance 0.9m above the floor.

There is no frictional force present here so, any type of losses will not occur here. Thus, the energy of the system when it is released from rest equal to the system just before m1 strikes the ground. So, energy is always conserved in ordinary conditions.

Conclusion:

Therefore, the energy is conserved in the absence of frictional force so, energy of the system when it is released from rest equal to the system just before m1 strikes the ground.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
A arrow of mass 0.01 kg moving horizontal suddenly strikes a block of wood of mass 8.6 kg that is suspended by a light string, like a pendulum. The arrow passes through the wood, then the wood swings upward and momentarily stops when the string is horizontal. The length of the string is 2.0 cm. How long did it take the arrow to go through the wood if the friction force between the bullies and the wood is 100 N?
As shown, a light string that does not stretch changes from horizontal to vertical as it passes over the edge of a table. The string connects m1, a 3.50-kg block originally at rest on the horizontal table at a height h = 1.20 m above the floor, to m2, a hanging 1.90-kg block originally a distance d = 0.900 m above the floor. Neither the surface of the table nor its edge exerts a force of kinetic friction. The blocks start to move from rest. The sliding block m1 is projected horizontally after reaching the edge of the table. The hanging block m2 stops without bouncing when it strikes the floor. Consider the two blocks plus the Earth as the system. (a) Find the speed at which m1 leaves the edge of the table. (b) Find the impact speed of m1 on the floor. (c) What is the shortest length of the string so that it does not go taut while m1 is in flight? (d) Is the energy of the system when it is released from rest equal to the energy of the system just before m1 strikes the ground? (e) Why or…
A 1.1 g pebble is stuck in a tread of a 0.74 m diameter automobile tire, held in place by static friction that can be at most 3.5 N . The car starts from rest and gradually accelerates on a straight road. How fast is the car moving when the pebble flies out of the tire tread?

Chapter 8 Solutions

Physics for Scientists and Engineers with Modern Physics

Ch. 8 - A crate of mass 10.0 kg is pulled up a rough...Ch. 8 - A 40.0-kg box initially at rest is pushed 5.00 m...Ch. 8 - Prob. 9PCh. 8 - As shown in Figure P8.10, a green bead of mass 25...Ch. 8 - At time ti, the kinetic energy of a particle is...Ch. 8 - A 1.50-kg object is held 1.20 m above a relaxed...Ch. 8 - Prob. 13PCh. 8 - An 80.0-kg skydiver jumps out of a balloon at an...Ch. 8 - You have spent a long day skiing and are tired....Ch. 8 - The electric motor of a model train accelerates...Ch. 8 - An energy-efficient lightbulb, taking in 28.0 W of...Ch. 8 - An older-model car accelerates from 0 to speed v...Ch. 8 - Prob. 19PCh. 8 - There is a 5K event coming up in your town. While...Ch. 8 - Prob. 21PCh. 8 - Energy is conventionally measured in Calories as...Ch. 8 - A block of mass m = 200 g is released from rest at...Ch. 8 - Prob. 24APCh. 8 - Prob. 25APCh. 8 - Review. As shown in Figure P8.26, a light string...Ch. 8 - Consider the blockspringsurface system in part (B)...Ch. 8 - Why is the following situation impossible? A...Ch. 8 - Jonathan is riding a bicycle and encounters a hill...Ch. 8 - Jonathan is riding a bicycle and encounters a hill...Ch. 8 - As the driver steps on the gas pedal, a car of...Ch. 8 - As it plows a parking lot, a snowplow pushes an...Ch. 8 - Prob. 33APCh. 8 - Prob. 34APCh. 8 - A horizontal spring attached to a wall has a force...Ch. 8 - Prob. 36APCh. 8 - Prob. 37APCh. 8 - Review. Why is the following situation impossible?...Ch. 8 - Prob. 39APCh. 8 - A pendulum, comprising a light string of length L...Ch. 8 - Prob. 41APCh. 8 - Prob. 42APCh. 8 - Prob. 43APCh. 8 - Starting from rest, a 64.0-kg person bungee jumps...Ch. 8 - Prob. 45CPCh. 8 - A uniform chain of length 8.00 m initially lies...Ch. 8 - Prob. 47CP
Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers with Modern ...
Physics
ISBN:9781337553292
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Text book image
College Physics
Physics
ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Impulse Derivation and Demonstration; Author: Flipping Physics;https://www.youtube.com/watch?v=9rwkTnTOB0s;License: Standard YouTube License, CC-BY