EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 4, Problem 65P

(a)

To determine

The reading of the scale when the elevator is ascending with the speed of 30m/s .

(a)

Expert Solution
Check Mark

Answer to Problem 65P

The reading indicated by the scale is 19.6N .

Explanation of Solution

Given:

The mass of the block is m=2.0kg .

Formula used:

The expression for the weight of the block is given by,

  w=mg

As the object is ascending at a constant speed, thus the reading of scale will be equal to the tension in the string and the expression for the tension of the string is given by,

  T=w

Calculation:

The weight of the block is calculated as,

  w=mg=(2kg)(9.81m/ s 2)=19.6N

The tension of the string is calculated as,

  T=w=19.6N

Conclusion:

Therefore, the reading indicated by the scale is 19.6N .

(b)

To determine

The reading of the scale when the elevator is descending with the speed of 30m/s .

(b)

Expert Solution
Check Mark

Answer to Problem 65P

The reading indicated by the scale is 19.6N .

Explanation of Solution

Given:

The mass of the block is m=2.0kg .

Formula used:

The expression for the weight of the block is given by,

  w=mg

As the object is descending at a constant speed, thus the reading of scale will be equal to the tension in the string and the expression for the tension of the string is given by,

  T=w

Calculation:

The weight of the block is calculated as,

  w=mg=(2kg)(9.81m/ s 2)=19.6N

The tension of the string is calculated as,

  T=w=19.6N

Conclusion:

Therefore, the reading indicated by the scale is 19.6N .

(c)

To determine

The reading of the scale when the elevator is ascending with the speed of 20m/s and is descending at the rate of 3.0m/s2 .

(c)

Expert Solution
Check Mark

Answer to Problem 65P

The reading indicated by the scale is 25.6N .

Explanation of Solution

Given:

The mass of the block is m=2.0kg .

Formula used:

Consider the upward acceleration of the block is a=3m/s2 .

The expression for the reading of the scale is given by,

  T=m(g+a)

Calculation:

The value of the reading of the scale is calculated as,

  T=m(g+a)=(2kg)(9.81m/ s 2+3m/ s 2)=25.6N

Conclusion:

Therefore, the reading indicated by the scale is 25.6N .

(d)

To determine

The reading of the scale for the time interval 0<t<9.0s .

(d)

Expert Solution
Check Mark

Answer to Problem 65P

The reading indicated by the scale for different time interval is 0<t<5=19.6N5t<9=14.6N

Explanation of Solution

Given:

The mass of the block is m=2.0kg .

Formula used:

The expression for the reading of the elevator during the first 5s is given by,

  w=mg

The block comes to rest in 4s from the speed of 10m/s , thus the acceleration is negative and is given by,

  v=v0at

The reading of the scale will be equal to the tension in the string and the expression for the tension in the string is given by,

  T=m(ga)

Calculation:

The value of the reading of the scale during the first 5s is given by,

  w=mg=(2kg)(9.81m/ s 2)=19.6N

The acceleration of the block after 4s and is given by,

  v=v0at0=10m/s(4s)aa=2.5m/s2

The reading of the scale is calculated as,

  T=m(ga)=(2kg)(9.81m/ s 22.5m/ s 2)=14.6N

Thus, the reading of the scale for different time interval is given by,

  0<t<5=19.6N5t<9=14.6N

Conclusion:

Therefore, the reading indicated by the scale for different time interval is 0<t<5=19.6N5t<9=14.6N

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