(¦ Indicates a review question, which means it requires only a basic understanding of the material to answer. Questions without tins designation typically require integrating or extending the concepts presented thus far.) . Six blocks with different masses, m , each start from rest at the top of smooth, frictionless inclines having length d and vertical height h and slide down. Rank the order, from greatest to smallest, of the Final kinetic energies of the masses when they reach the bottom of the inclines after having traveled their full lengths. If any of the situations yield the same kinetic energies, give them the same ranking, (a) m = 10 kg; A=lm; and d = 10 m (b) m = 10 kg; h = 1 m; and d = 5 m (c) m = 5 kg; h = 0.5 m; and d = 10 m (d) m = 1 kg; k = 2 m; and d = 5 m (e) m= 1 kg; k = 0.5 m; and d = 5 m (f) m = 15 kg; h = 0.75 m; and d = 7.5 m

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Answer 1

Textbook Question

Chapter 3, Problem 35Q

(¦ Indicates a review question, which means it requires only a basic understanding of the material to answer. Questions without tins designation typically require integrating or extending the concepts presented thus far.)

. Six blocks with different masses, m, each start from rest at the top of smooth, frictionless inclines having length d and vertical height h and slide down. Rank the order, from greatest to smallest, of the Final kinetic energies of the masses when they reach the bottom of the inclines after having traveled their full lengths. If any of the situations yield the same kinetic energies, give them the same ranking,
(a) m = 10 kg; A=lm; and d = 10 m
(b) m = 10 kg; h = 1 m; and d = 5 m
(c) m = 5 kg; h = 0.5 m; and d = 10 m
(d) m = 1 kg; k = 2 m; and d = 5 m
(e) m= 1 kg; k = 0.5 m; and d = 5 m
(f) m = 15 kg; h = 0.75 m; and d = 7.5 m

Expert Solution & Answer

To determine

The rank of each blocks according to the kinetic energy when they reach at the bottom.

Answer to Problem 35Q

The rank of all blocks is,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m E=10g JE=10g JE=2.5g JE=2g JE=0.5g JE=11.25g J Rank 2Rank 2Rank 3Rank 4Rank 5Rank 1.

Explanation of Solution

Given information:

The mass, and height and length of its position are given,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m

Concept Used:

The potential energy, E=mgh.

Calculation:

At the height, the velocity is zero. So, there will be no kinetic energy. The total energy will be the potential energy itself. At ground, the potential energy will be zero. The initial potential energy will completely convert into the kinetic energy. So, the kinetic energy of the block when it reaches ground is equal to the potential energy at the starting point. Let g be the acceleration due to gravity, the energy of each block can be represented as,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m E=10g JE=10g JE=2.5g JE=2g JE=0.5g JE=11.25g J

Thus, the rank of each block,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m E=10g JE=10g JE=2.5g JE=2g JE=0.5g JE=11.25g J Rank 2Rank 2Rank 3Rank 4Rank 5Rank 1.

Conclusion:

The rank of all blocks is,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m E=10g JE=10g JE=2.5g JE=2g JE=0.5g JE=11.25g J Rank 2Rank 2Rank 3Rank 4Rank 5Rank 1.

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Answer 2

Textbook Question

Chapter 3, Problem 35Q

(¦ Indicates a review question, which means it requires only a basic understanding of the material to answer. Questions without tins designation typically require integrating or extending the concepts presented thus far.)

. Six blocks with different masses, m, each start from rest at the top of smooth, frictionless inclines having length d and vertical height h and slide down. Rank the order, from greatest to smallest, of the Final kinetic energies of the masses when they reach the bottom of the inclines after having traveled their full lengths. If any of the situations yield the same kinetic energies, give them the same ranking,
(a) m = 10 kg; A=lm; and d = 10 m
(b) m = 10 kg; h = 1 m; and d = 5 m
(c) m = 5 kg; h = 0.5 m; and d = 10 m
(d) m = 1 kg; k = 2 m; and d = 5 m
(e) m= 1 kg; k = 0.5 m; and d = 5 m
(f) m = 15 kg; h = 0.75 m; and d = 7.5 m

Expert Solution & Answer

To determine

The rank of each blocks according to the kinetic energy when they reach at the bottom.

Answer to Problem 35Q

The rank of all blocks is,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m E=10g JE=10g JE=2.5g JE=2g JE=0.5g JE=11.25g J Rank 2Rank 2Rank 3Rank 4Rank 5Rank 1.

Explanation of Solution

Given information:

The mass, and height and length of its position are given,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m

Concept Used:

The potential energy, E=mgh.

Calculation:

At the height, the velocity is zero. So, there will be no kinetic energy. The total energy will be the potential energy itself. At ground, the potential energy will be zero. The initial potential energy will completely convert into the kinetic energy. So, the kinetic energy of the block when it reaches ground is equal to the potential energy at the starting point. Let g be the acceleration due to gravity, the energy of each block can be represented as,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m E=10g JE=10g JE=2.5g JE=2g JE=0.5g JE=11.25g J

Thus, the rank of each block,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m E=10g JE=10g JE=2.5g JE=2g JE=0.5g JE=11.25g J Rank 2Rank 2Rank 3Rank 4Rank 5Rank 1.

Conclusion:

The rank of all blocks is,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m E=10g JE=10g JE=2.5g JE=2g JE=0.5g JE=11.25g J Rank 2Rank 2Rank 3Rank 4Rank 5Rank 1.

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Answer 3

Textbook Question

Chapter 3, Problem 35Q

(¦ Indicates a review question, which means it requires only a basic understanding of the material to answer. Questions without tins designation typically require integrating or extending the concepts presented thus far.)

. Six blocks with different masses, m, each start from rest at the top of smooth, frictionless inclines having length d and vertical height h and slide down. Rank the order, from greatest to smallest, of the Final kinetic energies of the masses when they reach the bottom of the inclines after having traveled their full lengths. If any of the situations yield the same kinetic energies, give them the same ranking,
(a) m = 10 kg; A=lm; and d = 10 m
(b) m = 10 kg; h = 1 m; and d = 5 m
(c) m = 5 kg; h = 0.5 m; and d = 10 m
(d) m = 1 kg; k = 2 m; and d = 5 m
(e) m= 1 kg; k = 0.5 m; and d = 5 m
(f) m = 15 kg; h = 0.75 m; and d = 7.5 m

Expert Solution & Answer

To determine

The rank of each blocks according to the kinetic energy when they reach at the bottom.

Answer to Problem 35Q

The rank of all blocks is,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m E=10g JE=10g JE=2.5g JE=2g JE=0.5g JE=11.25g J Rank 2Rank 2Rank 3Rank 4Rank 5Rank 1.

Explanation of Solution

Given information:

The mass, and height and length of its position are given,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m

Concept Used:

The potential energy, E=mgh.

Calculation:

At the height, the velocity is zero. So, there will be no kinetic energy. The total energy will be the potential energy itself. At ground, the potential energy will be zero. The initial potential energy will completely convert into the kinetic energy. So, the kinetic energy of the block when it reaches ground is equal to the potential energy at the starting point. Let g be the acceleration due to gravity, the energy of each block can be represented as,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m E=10g JE=10g JE=2.5g JE=2g JE=0.5g JE=11.25g J

Thus, the rank of each block,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m E=10g JE=10g JE=2.5g JE=2g JE=0.5g JE=11.25g J Rank 2Rank 2Rank 3Rank 4Rank 5Rank 1.

Conclusion:

The rank of all blocks is,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m E=10g JE=10g JE=2.5g JE=2g JE=0.5g JE=11.25g J Rank 2Rank 2Rank 3Rank 4Rank 5Rank 1.

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Answer 4

Textbook Question

Chapter 3, Problem 35Q

(¦ Indicates a review question, which means it requires only a basic understanding of the material to answer. Questions without tins designation typically require integrating or extending the concepts presented thus far.)

. Six blocks with different masses, m, each start from rest at the top of smooth, frictionless inclines having length d and vertical height h and slide down. Rank the order, from greatest to smallest, of the Final kinetic energies of the masses when they reach the bottom of the inclines after having traveled their full lengths. If any of the situations yield the same kinetic energies, give them the same ranking,
(a) m = 10 kg; A=lm; and d = 10 m
(b) m = 10 kg; h = 1 m; and d = 5 m
(c) m = 5 kg; h = 0.5 m; and d = 10 m
(d) m = 1 kg; k = 2 m; and d = 5 m
(e) m= 1 kg; k = 0.5 m; and d = 5 m
(f) m = 15 kg; h = 0.75 m; and d = 7.5 m

Expert Solution & Answer

To determine

The rank of each blocks according to the kinetic energy when they reach at the bottom.

Answer to Problem 35Q

The rank of all blocks is,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m E=10g JE=10g JE=2.5g JE=2g JE=0.5g JE=11.25g J Rank 2Rank 2Rank 3Rank 4Rank 5Rank 1.

Explanation of Solution

Given information:

The mass, and height and length of its position are given,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m

Concept Used:

The potential energy, E=mgh.

Calculation:

At the height, the velocity is zero. So, there will be no kinetic energy. The total energy will be the potential energy itself. At ground, the potential energy will be zero. The initial potential energy will completely convert into the kinetic energy. So, the kinetic energy of the block when it reaches ground is equal to the potential energy at the starting point. Let g be the acceleration due to gravity, the energy of each block can be represented as,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m E=10g JE=10g JE=2.5g JE=2g JE=0.5g JE=11.25g J

Thus, the rank of each block,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m E=10g JE=10g JE=2.5g JE=2g JE=0.5g JE=11.25g J Rank 2Rank 2Rank 3Rank 4Rank 5Rank 1.

Conclusion:

The rank of all blocks is,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m E=10g JE=10g JE=2.5g JE=2g JE=0.5g JE=11.25g J Rank 2Rank 2Rank 3Rank 4Rank 5Rank 1.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

The rank of each blocks according to the kinetic energy when they reach at the bottom.

Answer to Problem 35Q

The rank of all blocks is,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m E=10g JE=10g JE=2.5g JE=2g JE=0.5g JE=11.25g J Rank 2Rank 2Rank 3Rank 4Rank 5Rank 1.

Explanation of Solution

Given information:

The mass, and height and length of its position are given,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m

Concept Used:

The potential energy, E=mgh.

Calculation:

At the height, the velocity is zero. So, there will be no kinetic energy. The total energy will be the potential energy itself. At ground, the potential energy will be zero. The initial potential energy will completely convert into the kinetic energy. So, the kinetic energy of the block when it reaches ground is equal to the potential energy at the starting point. Let g be the acceleration due to gravity, the energy of each block can be represented as,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m E=10g JE=10g JE=2.5g JE=2g JE=0.5g JE=11.25g J

Thus, the rank of each block,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m E=10g JE=10g JE=2.5g JE=2g JE=0.5g JE=11.25g J Rank 2Rank 2Rank 3Rank 4Rank 5Rank 1.

Conclusion:

The rank of all blocks is,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m E=10g JE=10g JE=2.5g JE=2g JE=0.5g JE=11.25g J Rank 2Rank 2Rank 3Rank 4Rank 5Rank 1.

Answer 8

Expert Solution & Answer

To determine

The rank of each blocks according to the kinetic energy when they reach at the bottom.

Answer to Problem 35Q

The rank of all blocks is,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m E=10g JE=10g JE=2.5g JE=2g JE=0.5g JE=11.25g J Rank 2Rank 2Rank 3Rank 4Rank 5Rank 1.

Explanation of Solution

Given information:

The mass, and height and length of its position are given,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m

Concept Used:

The potential energy, E=mgh.

Calculation:

At the height, the velocity is zero. So, there will be no kinetic energy. The total energy will be the potential energy itself. At ground, the potential energy will be zero. The initial potential energy will completely convert into the kinetic energy. So, the kinetic energy of the block when it reaches ground is equal to the potential energy at the starting point. Let g be the acceleration due to gravity, the energy of each block can be represented as,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m E=10g JE=10g JE=2.5g JE=2g JE=0.5g JE=11.25g J

Thus, the rank of each block,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m E=10g JE=10g JE=2.5g JE=2g JE=0.5g JE=11.25g J Rank 2Rank 2Rank 3Rank 4Rank 5Rank 1.

Conclusion:

The rank of all blocks is,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m E=10g JE=10g JE=2.5g JE=2g JE=0.5g JE=11.25g J Rank 2Rank 2Rank 3Rank 4Rank 5Rank 1.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

The rank of each blocks according to the kinetic energy when they reach at the bottom.

Answer 12

Answer to Problem 35Q

The rank of all blocks is,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m E=10g JE=10g JE=2.5g JE=2g JE=0.5g JE=11.25g J Rank 2Rank 2Rank 3Rank 4Rank 5Rank 1.

Answer 13

Explanation of Solution

Given information:

The mass, and height and length of its position are given,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m

Concept Used:

The potential energy, E=mgh.

Calculation:

At the height, the velocity is zero. So, there will be no kinetic energy. The total energy will be the potential energy itself. At ground, the potential energy will be zero. The initial potential energy will completely convert into the kinetic energy. So, the kinetic energy of the block when it reaches ground is equal to the potential energy at the starting point. Let g be the acceleration due to gravity, the energy of each block can be represented as,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m E=10g JE=10g JE=2.5g JE=2g JE=0.5g JE=11.25g J

Thus, the rank of each block,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m E=10g JE=10g JE=2.5g JE=2g JE=0.5g JE=11.25g J Rank 2Rank 2Rank 3Rank 4Rank 5Rank 1.

Conclusion:

The rank of all blocks is,

(a)m=10 kgh=1 md=10 m(b)m=10 kgh=1 md=5 m(c)m=5 kgh=0.5 md=10 m(d)m=1 kgh=2 md=5 m(e)m=1 kgh=0.5 md=5 m(f)m=15 kgh=0.75 md=7.5 m E=10g JE=10g JE=2.5g JE=2g JE=0.5g JE=11.25g J Rank 2Rank 2Rank 3Rank 4Rank 5Rank 1.

Answer 14

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