Question

The position of a 4 kg object changes from (2i -5j) to 4i+2j. Constant resultant force acting on is equal to (6i-5j) N. is the speed for metres per second what is its kinetic energy at its final position

Answer 1

Mass of the object, m = 4 kg

Displacement of the object, r = (4i+2j) – (2i -5j) = (3i + 7j ) in m

Force applied, F = (6i-5j) N

Work done, W = F.r = (6i-5j . (3i + 7j ) = 18 – 35 = -17 J

Work done on the object, W = 17 J

Using work energy theorem,

Kinetic energy = work done

Kinetic energy, E = W = 17 J

Answer; kinetic energy at its final position, E = 17 J

Also, E = ½ mv2 where v is the speed

V2 = 2E/m = 2 x 17 /4 = 8.5

V = 2.92 m/s

Answer: speed of the object at the final position, v = 2.92 m/s