A sphere of mass 0.5 kg rests at one corner of a cube. The cube is moved with velocity `v=8hati-2t^(2)hatj` where t is time in seconds. Find the magnitude of force by sphere on the cube at t=1 s is (figure shown vertical plane of the cube `g=10m//s^(2)`)

Explanation

`a=(dv)/(dt)=8hati-4thatj t=1 a=8hati-4hatj`
|force on sphere by cube|=|force on cube by sphere|

`F_("sphere")rArr-8mhati+6mhatj=-4hati+3hatj`
`|F|rArr5N`