A particle is projected with a speed {v_ | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$\mathrm{N}-\mathrm{mg}=\frac{\mathrm{mv}^{2}}{\mathrm{R}}=\mathrm{mg}$

$\mathrm{f}_{\mathrm{k}}=\mu \mathrm{N}$

$\mathrm{f}_{\mathrm{k}}=2 \mu

Vedclass · Accepted Answer

$\sqrt 2 g\, 
warrow $

Vedclass · Answer

$\mathrm{N}-\mathrm{mg}=\frac{\mathrm{mv}^{2}}{\mathrm{R}}=\mathrm{mg}$

$\mathrm{f}_{\mathrm{k}}=\mu \mathrm{N}$

$\mathrm{f}_{\mathrm{k}}=2 \mu \mathrm{mg}$

$\mathrm{ma}_{\mathrm{T}}=2 \mu \mathrm{mg}$

$a_{T}=2 \mu g(\leftarrow)$

$\mathrm{a}_{\mathrm{N}}=\frac{\mathrm{v}^{2}}{\mathrm{R}}=(\uparrow)$

$a=\sqrt{a_T^2+a_N^2}$

$a=\sqrt{2} g$ $ 
warrow $

A particle is projected with a speed ${v_0} = \sqrt {gR} $ . The coefficient of friction between the particle and the hemispherical plane is $\mu = 0.5$ . Then, the initial acceleration of the particle is

Similar Questions

A block of mass $m$ is placed on a surface with a vertical cross section given by $y = \frac{{{x^3}}}{6}$ If the coefficient of friction is $0.5$,the maximum height above the ground at which the block can be placed without slipping is:

A block of mass $10\, kg$ starts sliding on a surface with an initial velocity of $9.8\, ms ^{-1}$. The coefficient of friction between the surface and bock is $0.5$. The distance covered by the block before coming to rest is: [use $g =9.8\, ms ^{-2}$ ].........$m$

Consider a car moving on a straight road with a speed of $100\, m/s$. The distance at which car can be stopped, is ........ $m$. $[\mu_k = 0.5]$

A block of mass $10\, kg$ is placed on a rough horizontal surface having coefficient of friction $\,\mu = 0.5$. If a horizontal force of $100\, N$ is acting on it, then acceleration of the block will be ....... $m/s^2$

Write relation between coefficient of static friction, kinetic friction and rolling friction.