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The staples in a stapler are pulled into position by a spring with a spring constant of 50 N/m.  When the stapler is open, the spring is 8 cm long and relaxed.  When you close the stapler, the spring stretches to 15 cm long.  How much force (in Newton) does it exert on the staples?
 

Option: 1

3.5


Option: 2

5


Option: 3

2.5


Option: 4

1.5


Answers (1)

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As we learned

1.   Spring Force:-

  • Spring force is also called restoring force.

  • F= -Kx 

           where k is the spring constant and its unit is N/m and x is net elongation or compression in the spring.

  • Here -ve sign is because the force exerted by the spring is always in the opposite direction to the displacement. 

Given,

k = 50\; N/m

 l_{0}= 8 cm = 0.08 m

l = 15 cm = 0.15

Here \; x=l-l_{0}=.15-.08=.07m \Rightarrow it\; is\; the\; net \; elongation.

F_{s}=-kx=-(50 \times .07) N= -3.5 N/m \Rightarrow here the negative means that the spring pulls inward

because it is longer than normal.

Posted by

HARSH KANKARIA

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