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# Spring Potential Energy Problem

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 as shown in the figure. System is in equilibrium. Assuming that all surfaces are smooth, the potential energy stored on the spring is:

A)$\frac{mg^2tan^2\theta}{2k}$ B)$\frac{m^2gtan^2\theta}{2k}$ C)$\frac{m^2g^2tan^2\theta}{2k}$ D)$\frac{m^2g^2tan^2\theta}{k}$

Note by Shubham Srivastava
4 years, 4 months ago

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Make free body diagram of forces on both the blocks one by one. For wedge of mass M the forces acting along the floor are N sinθ (normal reaction from block of mass m) and kx. Next, make free body diagram of block of mass m. Forces acting on it perpendicular to the floor are N cosθ and mg. Solving these equations

Nsinθ = kx,

Ncosθ = mg,

We get the value of x, hence value of potential energy can be evaluated. · 4 years, 4 months ago

c · 4 years, 4 months ago

Solution???? · 4 years, 4 months ago

Start with determining the state of spring at equilibrium. Should it be compressed or stretched to keep the system at equilibrium? · 4 years, 4 months ago

make forces acting on M, weight of m & a force F which is acts opposite to spring force to keep system is in equilibrium. then \Tan\theta\ = F\mg. from this find out F and substituted in P.E = F*2\ 2K get the ans... (C) is correct · 4 years, 3 months ago

My attitude towards the problem is this: The weight, mg, of the rod will have 2 components- mgsin0 and mgcos0. The component, mgcos0, can be again resolved into two vectors having a horizontal component, mgcos0sin0. For equilibrium, the spring force should be equal to horizontal force. Therefore, mgcos0sin0= kx. We can now substitute value of x from this equation in the equation for spring potential energy, i.e., 0.5kx^2. · 4 years, 4 months ago

I fear your reasoning is not accurate. You are making components of components. This is probably wrong way and you will get the wrong answer I think. · 4 years, 4 months ago

mgcos0 (0 to be read as theta) is not a component but the normal reaction of m on M... So there are two forces on M....one is the spring force kx which is horizontal and towards right...and the other is the normal reaction of m on M which towards left, and downwards making an angle of (ninty - theta) with the horizontal...So, on resolving mgcos0 in the horizontal left direction, we get mgcos0sin0...and for M to remain in equillibrium, kx should be equal to mgcos0sin0...

so,

mgcos0sin0 = kx

x=(mgcos0sin0)/k

the potential energy stored in spring = 1/2(k . x^2)..

solvong further will give the answer... So the explanation of Shubham S is accurate.... · 4 years, 2 months ago

BY DIMENSIONAL ANALYSIS IT CAN ONLY BE C OR D MOSTLY C · 4 years, 4 months ago

Mind keeping Caps Lock off. :P · 4 years, 4 months ago

how can u say it is mostly c)? · 4 years, 3 months ago

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