A body of mass m kg is ascending on a smooth inclined plane of inclination `theta(sintheta=(1)/(x))` with constant acceleration of a `m//s^(2).` The final velocity of the body is v `m//s^(2).` The work done by the body during this motion is (Initial velocity of the body = 0)

Options

A. `(1)/(2)mv^(2)(g+xa)`
B. `(mv^(2))/(2)((g)/(2)+a)`
C. `(2mv^(2)x)/(a)(a+gx)`
D. `(mv^(2))/(2ax)(g+xa)`

Answer

D

Explanation

`Sin theta=(1)/(x)`
From free body diagram of the body
F-mg sin`theta` =ma
F=m(g sin `theta`+a)
`m((g)/(x)+a)`……….(1)
Displacement of the body
till its velocity reaches v
`v^(2)=0+2asimpliess(v^(2))/(2a)`
Now,work done=F s cos `0^(@)`
`2=(m)/(x)(g+ax)xx(v^(2))/(2a)=(mv^(2))/(2ax)(g+ax)`