Physics, asked by Spartanyashwant4842, 1 year ago

A wedge of mass M fitted with a spring of stiffness k is kept on a smooth horizontal surface. A rod of mass m is kept on the wedge .System is in equilibrium and at rest Assuming that all surfaces are smooth, the potential energy stored in the spring is(theta is the angle between the inclined plane of Wedge M and horizontal surface ) Ans- m^2 g^2 tan^2 theta/2K

Answers

Answered by Anonymous
33
Force due to spring, Fs = kx

Force due to gravitation, Fg = mg sinϴ

Fs = Fg

kx = mg sinϴ

x = mg sinϴ/k

Potential Energy stored in the spring, U

= kx^2/2 = k(mg sinθ)^2/2k = m^2g^2sinθ^2/2k

piyush8700: wrong answer
Answered by ankush4b725
86

Answer:

reffer

above

answer

cheers

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