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)\[\frac{m^2g^2tan^2\theta}{k}\]

No vote yet

8 votes

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestMake 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.

Log in to reply

c

Log in to reply

Solution????

Log in to reply

kx=mgtan@

x=mgtan@/2k

1/2 kx^2 = m^2g^2tan^2/2k

Log in to reply

Start with determining the state of spring at equilibrium. Should it be compressed or stretched to keep the system at equilibrium?

Log in to reply

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

Log in to reply

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.

Log in to reply

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.

Log in to reply

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....

Log in to reply

BY DIMENSIONAL ANALYSIS IT CAN ONLY BE C OR D MOSTLY C

Log in to reply

Mind keeping Caps Lock off. :P

Log in to reply

how can u say it is mostly c)?

Log in to reply