A uniform rod of length L and mass m is placed on rough horizontal surface-If coefficient of friction is - kx- where k is constant and - exist for 0 - friction is-A- kmgL23B- 2kmgL23C- kmgL26D- kmgL29

The correct option is C kmgL26 df=μ dN = kxm/L dxg f=∫0LkxmLg nbsp;dxwork nbsp;done=force×diplacementdw=(kxmLdx.g) nbsp;(L x)W=∫0LkmgLx(L x)dx nbsp; ...

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Byju's Answer

Standard XII

Physics

Newton's Laws for System of Particles

A uniform rod...

Question

A uniform rod of length L and mass m is placed on rough horizontal surface.If coefficient of friction is $μ = k x$ ,where k is constant and $μ$ exist for $0 \leq x \leq L$ (i.e for $x > L$ friction is zero.). If rod moves towards the right, magnitude of work done by friction is:

\frac{k m g L^{2}}{9}

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\frac{k m g L^{2}}{6}

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\frac{k m g L^{2}}{3}

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\frac{2 k m g L^{2}}{3}

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Solution

The correct option is B $\frac{k m g L^{2}}{6}$

$d f = μ d N = k x \frac{m}{L} d x g$
$f = \int_{0}^{L} k x \frac{m}{L} g d x w o r k d o n e = f o r c e \times d i p l a c e m e n t d w = (k x \frac{m}{L} d x . g) (L - x) W = \int_{0}^{L} \frac{k m g}{L} x (L - x) d x = \frac{k m g L^{2}}{6}$

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A uniform rod of length L and mass m is placed on rough horizontal surface.If coefficient of friction is μ=kx,where k is constant and μ exist for 0≤ x≤ L (i.e for x>L friction is zero.). If rod moves towards the right, magnitude of work done by friction is:

The correct option is B kmgL26 df=μdN=kxmLdxg f=∫L0kxmLg dxwork done=force×diplacementdw=(kxmLdx.g) (L−x)W=∫L0kmgLx(L−x)dx =kmgL26

A uniform rod of length L and mass m is placed on rough horizontal surface.If coefficient of friction is $μ = k x$ ,where k is constant and $μ$ exist for $0 \leq x \leq L$ (i.e for $x > L$ friction is zero.). If rod moves towards the right, magnitude of work done by friction is:

The correct option is B $\frac{k m g L^{2}}{6}$

$d f = μ d N = k x \frac{m}{L} d x g$
$f = \int_{0}^{L} k x \frac{m}{L} g d x w o r k d o n e = f o r c e \times d i p l a c e m e n t d w = (k x \frac{m}{L} d x . g) (L - x) W = \int_{0}^{L} \frac{k m g}{L} x (L - x) d x = \frac{k m g L^{2}}{6}$