Two block of masses 'M' and 'm' are placed on one another on a smoothe horizontal surface as shown in the figure.
The force 'F' is acting on the mass 'M' horizontally during time intervel 't' Assuming no relative sliding between the blocks, the work done by friction on the blocks is . . . . .

Options

A. `(Ft)/(2(M + m))`
B. `(M + m)/(mt^(2))`
C. `(mF^(2)t^(2))/(2(M+m)^(2))`
D. `(F^(2) t^(2))/((M +m))`

Answer

C

Explanation

Acceleration of block `= (F)/((M + m))`
So, friction acting over upper block , f
`ma= (mF)/((M + m))`
Distances moved in time t `= -(1)/(2) at^(2)`
` = (1)/(2) ((F)/(M = m))t^(2)`
Work done by friction ` f = xxs`
`= ((mF)/(M + m)) xx (1)/(2) (Ft^(2))/((M + m))`
`= (mF^(2)t^(2))/(2(M+m)^(2))`