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A load of 36 N attached to a spring hanging
vertically stretches the spring 6.5 cm. The
spring is now placed horizontally on a table
and stretched 10 cm.
What force is required to stretch it by this
amount?
Answer in units of N.

Respuesta :

Answer:

The required force is 55.38 N.

Explanation:

First we have to find the spring constant K;

                         k = [tex]\frac{mg}{d}[/tex] = [tex]\frac{36}{0.065}[/tex] = 553.846 N/m

As,

                         F= -K[tex]x[/tex]

                   or, F=  - 553.846 × 0.10

                   or, F=  -55.3846 N

Hence,

                  [tex]F_{needed}[/tex] = - [tex]F_{spring}[/tex]

            ∴  [tex]F_{needed}[/tex] = 55.38 N

The required force  to stretch it by this  amount is 55.38 N.  

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