A block of mass m is stationary with respect to a rough wedge as shown in figure. Starting from rest in time t, (m = 1 kg, `theta=30^(@), a = 2 m//s^(2)`, t = 4 s) work done on block :

Answer

`(A to t, B to p, C to s, D to q)`

Explanation

In t=4 s, v=at`=8ms^(-1)`
and`" " s=(1)/(2)at^(2)`=16m
`KE=(1)/(2)mv^(2)=32J`
From work-energy theorem,
Work done by all the forces`=DeltaKE=32J`
Work done by gravity,`W_(g)=-mgh=-(1)(10)(16)=-160J`
Writing equation of motion, we have, `Sigma F_(y)`=ma

`N cos30^(@)+f sin30^(@)-10`=ma=2
or `" " sqrt3N +f=24" ".........(i)`
`SigmaF_(x)=0`
`:. " " N sin 30^(@)=f cos 30^(@)`
or `" " N=sqrt3f " " .....(ii)`
Solving Eqs. (i) and (ii), we have
f=6N and `N=6sqrt3 N`
Now, `W_(N)=(N cos theta)(s)`
`=(6sqrt3)((sqrt3)/(2))(16)=144J`
`W_(f)=(f sin theta)(s)=(6)(1)/(2)(16)=48J`
Work done by all the forces
`W=W_(g)+W_(N)+W_(F)=-160+144+48=32J`