A sphere of mass `1 kg` rests at one corner of a cube. The cube is moved with a velocity `v =(8 t hat(i) - 2t^(2)) hat(j)`, where `t`is time in second . The force by sphere on the cube at `t = 1 s` is `(g = 10 m//s^(-2))` [figure shown vertical plane of the cube]

Options

A. `8 N`
B. `10 N`
C. `20 N`
D. `6 N`

Answer

B

Explanation

`a = (dv)/(dt) = (8hati - 4t hatj)`
At `1 s F_("net") = ma = (1) (8hati - 4hatj) = (8hati - 4hatj)`
`= W+F`
where `F =`force on cube
`:. F = (8hati - 4hatj) - w`
` = (8hati - 4hatj) - (-10hatj)`
`= (8hati + 6hatj)`
or `|F|= sqrt((8)^(2) + (6)^(2)`
`= 10 N`