A force of `20 N` is applied on upper block as shown in the figure. Te total work done by frictional during the time interval in which the upper block has a displacement of `15 m` with respect to the ground is (take `g = 10 m//s^(2)` )

Options

A. `50 J`
B. `-50 J`
C. `75 J`
D. `- 75 J`

Answer

B

Explanation

Block and plank both slide
so, `a_("Block") = (20 -5)/(2) = (15)/(2)`
`a_("plank") = (5)/(2)`
`S = ut + (1)/(2)at^(2)`
`15 = 0 + (1)/(2) (15)/(2)t^(2)`
`rArr t = 2 sec`
Desplacement of plane `= 0 + (1)/(2) (5)/(2)2^(2)`
So, displacement of block with respect to plank `= 15 - 5 = 10m`
`W_(f) = - 10 xx 20 xx 1//4 = -50J`