An ideal spring is hung vrrtically from a celling when a 2 kg maddhanfs at rest from it
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Mathematically, Hooke's law states that F = -kx (The negative sign indicates that the force exerted by the spring is in direct opposition to the direction of displacement)
where
x is the displacement of the end of the spring from its equilibrium position = 0.06 m
F is the restoring force exerted by the material in N or kgm/s2);
and k is a constant called the rate or spring constant in "N·m-1"
F=mg =2*9.81 = k*0.06
or spring constant "k" = 327 N/m
Now since spring is further streched by 10 cm or say 0.1 m (the energy stored in spring is work done upon spring by streching it which would be
W = 1/2 k x^2
= 1/2 * 327 *(0.1)^2
= 1.635 Nm or Joule
Hope this helpe
where
x is the displacement of the end of the spring from its equilibrium position = 0.06 m
F is the restoring force exerted by the material in N or kgm/s2);
and k is a constant called the rate or spring constant in "N·m-1"
F=mg =2*9.81 = k*0.06
or spring constant "k" = 327 N/m
Now since spring is further streched by 10 cm or say 0.1 m (the energy stored in spring is work done upon spring by streching it which would be
W = 1/2 k x^2
= 1/2 * 327 *(0.1)^2
= 1.635 Nm or Joule
Hope this helpe
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