A person pushes horizontally on a heavy box and slides it across the level floor at constant velocity. The person pushes with a
60.0 N force for the first 7.09 m, at which time he begins to tire. The force he exerts then starts to decrease linearly from 60.0 N to 0.00 N across the remaining 7.09 m. How much total work did the person do on the box?
1 answer:
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Answer:
The ratio of the energy stored by spring #1 to that stored by spring #2 is 2:1
Explanation:
Let the weight that is hooked to two springs be w.
Spring#1:
Force constant= k
let x1 be the extension in spring#1
Therefore by balancing the forces, we get
Spring force= weight
⇒k·x1=w
⇒x1=w/k
Energy stored in a spring is given by where k is the force constant and x is the extension in spring.
Therefore Energy stored in spring#1 is,
⇒
⇒
Spring #2:
Force constant= 2k
let x2 be the extension in spring#2
Therefore by balancing the forces, we get
Spring force= weight
⇒2k·x2=w
⇒x2=w/2k
Therefore Energy stored in spring#2 is,
⇒
⇒
∴The ratio of the energy stored by spring #1 to that stored by spring #2 is 2:1