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A particle of mass 2Kg revovles in a circular path on X-Y plane with a constant speed of 5 m/s along the path x2+y2-25=0.The force(in N) acting on the particle at the position (-3m,4m) will be_______________________
Dear sindhuja
radius of the circle is =5
since it is moving with constant speed so there is no tangential acceleration
only centripetal acceleraton will act towards the center of circle .
centripetal force =mv2/r
=2*25/5 =10 N
tan θ =|4/-3|
=4/3
so force =10(cosθ i -sinθj)
=6i-8j
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