Jackson ran in a marathon for 3 hours. The marathon was 185 miles long. How fast was the runner running?
1 answer:
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Answer:
a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve is
.
Explanation:
a) The work, measured in joules, is a physical variable represented by the following integral:
Where
,
- Initial and final position, respectively, measured in meters.
- Force as a function of position, measured in newtons.
Given that and the fact that
when
, the spring constant (
), measured in newtons per meter, is:
Now, the work function is obtained:
The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.
b) Let be . The area of the region enclosed by one loop of the curve is given by the following integral:
By using trigonometrical identities, the integral is further simplified:
The area of the region enclosed by one loop of the curve is
.