In the figure shown, upper block is given a velocity of `6 m//s` and lower block. `3 m//s`. When relative motion between them is stopped

Options

A. Work done by friction on upper block is –10 J
B. Work done by friction on lower block is + 10 J
C. Net work done by friction on the system is zero
D. Net work done by friction on the system is – 3 J

Answer

A, D

Explanation

From conservation of linear momentum,
`(1+2) v = (6 xx 1) + (2 xx 3)`
`v = 4m//s` (of both the blocks)
From work energy theorem i.e., `W_("Total") = Delta KE`
On 1 kg block,`W_(f) = 1/2 xx 1 (4^2 - 6^2) = -10 J`
On 2 kg block, `W_(f) = 1/2 xx 2(4^2 - 3^2) = +7 J " "` Net work done by friction is –3J.