A force of F=0.5 N is applied on lower block as shown in figure. The work done by lower block on upper block for a displacement of 3 m of the upper block with respect to ground is (Take, g=10 `ms^(-2)`)

Options

A. `-0.5 J`
B. 0.5 J
C. 2 J
D. `-2 J`

Answer

B

Explanation

(b) Maximum acceleration of 1 kg block may be
`a_(max)=mug=1ms^(-2)`
Common acceleration, without relative motion between two blocks may be,
`a=(0.5)/(3)ms^(-2)`
Since, `a lt a_(max)`
There will be no relative motion and blocks will move with acceleration `(0.5)/(3)ms^(-2)`.
Force of friction by lower block on upper block,
`f=ma=(1)((0.5)/(3))=(1)/(6)N("towards right")`
`:. " " W=fxxs=0.5J`