Question

a particle is placed at the point A of a frictionless track ABC . where point a is at a height of 1m point b at 0.5m and point c at little less than point a . it is pushed slightly towards right.find its speed when it reaches the point B take g = 10m/s^2

Answer 1

Heya frnd here's your answer

the potential energy at A is mgh=m×10×1=10m
(where m is mass)

kinetic energy at A is 0

mechanical energy at A is 10m+0=10m00m

potential energy at B = mgh = m×10×.5=5m

kinetic energy at B is 10m-5m=5m

hence velocity (or speed)
1/2mv^2=5m
1/2v^2=5
v^2=5×2
:.v=
$\sqrt{10}$
Hope this helps you