A mass m slips along wall of a semispherical surface of radius R . velocity at bottom of surface is

English

Gujarati

Hindi

5.Work, Energy, Power and Collision

normal

A mass $m$ slips along the wall of a semispherical surface of radius $R$. The velocity at the bottom of the surface is

A

$\sqrt {Rg} $

B

$\sqrt {2Rg} $

C

$2\sqrt {\pi Rg} $

D

$\sqrt {\pi Rg} $

Solution

(b) By applying law of conservation of energy

$mgR = \frac{1}{2}m{v^2} \Rightarrow v = \sqrt {2Rg} $

Standard 11

Physics

Share

Similar Questions

A uniform chain of length $L$ and mass $M$ is lying on a smooth table and one third of its length is hanging vertically down over the edge of the table. If $g$ is acceleration due to gravity, work required to pull the hanging part on to the table is

normal

A body of mass $m$ is moving in a circle of radius $r$ with a constant speed $v$. The force on the body is $\frac{mv^2}{r}$ and is directed towards the centre. What is the work done by this force in moving the body over half the circumference of the circle

normal

$A$ ball is dropped from height $5m$. The time after which ball stops rebounding if coefficient of restitution between ball and ground $e = 1/2$, is ……………… $\mathrm{sec}$

normal

The force acting on a body moving along $x-$ axis varies with the position of the particle as shown in the figure. The body is in stable equilibrium at

normal

Two identical $5\,\,kg.$ blocks are moving with same speed of $2\,\,m/s$ towards each other along a frictionless horizontal surface. The two blocks collide, stick together and come to rest. Consider to two blocks as a system, the work done on the system by the external forces will be ………….. $\mathrm{Joule}$

normal