A block lying on a rough horizontal surfaces as show in figure . Its mass is M and it is dispalced through a distance d by a force `vec(F )` acting at an angle `theta ` with horizontal . If `mu` is the coefficient of friction between a block and the surface and mass of block is M . then work done is `...........`

Options

A. `W = [ F(sin theta + mu cos theta) - mu Mg] d`
B. `W = [F (cos theta+musin theta) - mu Mg] d`
C. `W = [F sin theta +mu cos theta) +muMg] d`
D. `W=[F (cos theta - mu sin theta) + mu Mg] d `

Answer

B

Explanation

`W = [F (cos theta + mu sin theta) - mu Mg]d`
FBD of a block is shown in figure below :

As there is no displacement in Y - direction ,
`N + F in theta = Mg = 0 `
` :. N + F sin theta = Mg`
` :. N = Mg - F sin theta......(i)`
Force responsible for displacement in X direction ,
`F_(1) = F cos theta - f ` where , f = frictional force = `muN`
` = F co theta - mu N `
` :. F_(1) = F cos theta - mu (Mg - F sin theta)`
( From eq.(1) )`.....(ii)`
Now , work `W = F_(1)d`
` = [ F cos theta - mu Mg + Fmu sin theta ] d `
[ From wq. (2)]
` = [ F cos theta +F mu sin theta - mu Mg ] d `
` :. W = [F(cos theta +mu sin theta) - mu Mg ] d `