A small object of mass of 100 g moves in a circular path. At a given instant velocity of the object is 10 i ^ m/s and acceleration is ( 20 i ^ + 10 j ^ ) m / s 2 . At this instant of time, rate of change of kinetic energy (in kg m 2 s - 3 ) of the object is

Step-by-step solution

Given, mass of object m = 100 g = 100 × 10 - 3 k g Velocity of object v = 10 i ^ m / s Acceleration of object a = ( 20 i ^ + 10 j ^ ) m / s 2 We know that, d ( K E ) d t = F . v = m a . v [ ∵ K . E . = 1 2 m v 2 ] = ( 100 × 10 - 3 ) ( 20 i ^ + 10 j ^ ). ( 10 i ^ ) = 100 × 10 - 3 × 200 = 100 × 200 1000 = 20 kg m 2 / s 3