Answer to Question #174460 in Mechanics | Relativity for Ryan Camilleri

Question

A particle P of mass 5kg is suspended from a fixed point O by a light inextensible string of length 1m. The particle is projected from its lowest positon at the point A, with a horizontal speed of 4 ms-1. When angle AOP = 600. Find: a)The speed of P b)The tension in the string.

Answer 1

Given:

mass (m)=5kg

length(l)=1m

angle(θ)=60⁰

velocity(v)=4m\s

solution:

so according to problem we have to find tention(T) and speed (s) at 60^o or point B .

a)from the figure we can calculate speed s by,

energy conservation

${1\over2}mv^2={1\over2}ms^2+mg(l)(1-cos\theta)$

${1\over2}(4)(5)^2={1\over2}(5)s^2+(5)(10)(1)({1\over2})$

$\boxed{s=\sqrt10m/s}$

b)using this we can calculate Tension by formula

$\boxed{T={mv^2\over l}+mgcos\theta}$

$T={5(10)\over 1}+(5)(10)({1\over2})$

$\boxed{T=75N}$