(a)
To Find: The work done needs to be determined in compressing the spring the first half inch.
(a)
Answer to Problem 36E
The work done in compressing the spring the first half inch is
Explanation of Solution
Given information: A 10,000-lb force is compressing a spring from its natural length of 12 inches to a length of 11 inches.
Formula used: The Hooke’s law is used which says that when a spring is compressed or stretched x units from its natural length then the force it takes is a constant times x and in symbols it is as
Calculation:
Step 1:
When a 10,000-lb force is compressing a spring from its natural length of 12 inches to a length of 11 inches, then by Hooke’s law one has as:
Step 2:
Now further when spring is compressed first half inch then the force required is evaluated as:
Hence the work done in compressing the spring to first half inch is
(b)
To Find: The work done needs to be determined in compressing the spring the second half inch.
(b)
Answer to Problem 36E
The work done in compressing the spring the second half inch is
Explanation of Solution
Given information: A 10,000-lb force is compressing a spring from its natural length of 12 inches to a length of 11 inches.
Formula used: The Hooke’s law is used which says that when a spring is compressed or stretched x units from its natural length then the force it takes is a constant times x and in symbols it is as
Calculation:
Step 1:
When a 10,000-lb force is compressing a spring from its natural length of 12 inches to a length of 11 inches, then by Hooke’s law one has as:
Step 2:
Now further when spring is compressed second half inch then the force required is evaluated as:
Hence the work done in compressing the spring to second half inch is
Chapter 7 Solutions
CALCULUS-W/XL ACCESS
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