Question

As a 2.0-kg object moves from (11i-5j)m to (7i-8j)m, the constant resultant force acting on it is equal to (4i-3j)N. If the speed of the object at the initial position is 4.0 m/s, what is its kinetic energy at its final position?

Answer 1

m Mass of object = 2.0 kg Force acting on object = F = (42 - 3 f) N 11111 (432443,2 = 116+9 = 125 = 5N The deject moved from ă, to i - And 2 = (11 ê - 59)m x2 = (4 î - 8 ) m Displacement - n=,- =(11î - 5f) - (79-89) 3 61-7)Ê + (-5 +8) has (4 î +39) m 15= dius?+13)2 = 125 = 5 m Initial speed of object = 4.0 m/s = a (say) By, Newton's second law of motion I=ma 5 = 2 (a) a) a = 5 = 2.5 m/s² And we will use Newton's equation of motion v²_u²=2as v= final speed Sa displacement v2_4²= 2 as v2-1432 = (2)(25)(5) v2 - 16 - 25 12- 25 +16 =) v2-41 = v= 6. 40 mis final speed of object = v= 6:40 m/s

Kinetic Energy at its final position a / mn2 (K.E.) final a 4 x 2x 41 (K.E.)c = 41 J