Question

A particle of mass 1g executes an oscillatory motion on a concave surface of spherical dish of radius 2m placed on the horizontal plane. If the motion of the particle begins from a point on the dish at the height of 1cm from the horizontal plane and coefficient of friction is 0.01, find the total distance covered by the particle before it comes to rest.

Answer 1

Answer:

$\huge\fbox{☘Answer}$

Loss in mechanical energy = work done against friction 

mgh=μmgcosθ×s

s=μgcosθh=0.01×10×22−h1×10−2=1.005m≈1m

Answer 2

Given,

mass = 1g

Radius = 2m

height = 1 cm.

u = 0.01

Potential energy of particle = mgh .

Solñ,

Work done against the friction will be-

$\bold \orange{ \: \: w = {uR} \times d = umg \times d}$

As Potential energy is spent in doing the work.

$\sf {u mg \times d = mmgh} \\ \sf{d \: = \frac{h}{u} = \frac{1}{0.01} = 100cm = 1m}$

$\sf \green{Therefore,}$

1m is the distance covered by the particle before it comes to rest.