Question

1) An Object of Mass 2 Kg
Increases in speed from 2 m/s to 4
m/s in 3s. What was the total work
performed on the object during this
time interval?

Answer 1

Answer:

here a =v-u/t= 4-2/3=2/3

F=ma=4/3

S=v^2-u^2/2a

S= (16-4)×3/4=9

w=F×d= 4/3×9=12 Nm