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 placed on a rough horizontal surface is pulled by a horizontal force $F$. Let $f$ be the force applied by the rough surface on the block. Plot a graph of $f$ versus $F$.

A conveyor belt is moving at a constant speed of $2\, m s^{-1}$. A box is gently dropped on it. The coefficient of friction between them is $\mu = 0.5.$ The distance that the box will move relative to belt before coming to rest on it, taking $g = 10\, m s^{-2},$ is ........... $m$

A horizontal force of $4\,N$ is needed to keep a block of mass $0.5\, kg$ sliding on a horizontal surface with a constant speed. The coefficient of sliding friction must be :- $[g = 10\, m/s^2]$

A block of mass $m$ slides along a floor while a force of magnitude $F$ is applied to it at an angle $\theta$ as shown in figure. The coefficient of kinetic friction is $\mu_{ K }$. Then, the block's acceleration $'a'$ is given by: ($g$ is acceleration due to gravity)

A heavy box of mass $50 \mathrm{~kg}$ is moving on a horizontal surface. If co-efficient of kinetic friction between the box and horizontal surface is $0.3$ then force of kinetic friction is :