A block of mass M is pulled along a horizontal surface by applying a force at angle `theta` with the horizontal. The friction coefficient between the block and the surfasce is `mu`. If the block travels at a uniform velocity, find the work donen by this applied force during a displacement d of the blcok.

Options

A. ` ( mu mg d ) / ( cos theta + mu sin theta ) `
B. ` (mu mg d cos theta ) /( cos theta + mu sin theta ) `
C. ` ( mu mg d sin theta ) /( cos theta + mu sin theta ) `
D. ` ( mu mg d cos theta )/( cos theta - mu sin theta ) `

Answer

B

Explanation

Because the block moves with a uniform velocity, the resultant force is zero. Resolving F into horizontl component F ` cos theta ` and vertical component ` F sin theta ` , we get

` R + F sin theta = mg or R = mg - F sin theta `
Also ` f = mu R = mu ( mg - F sin theta ) `
But ` F cos theta = f `
or ` F cos theta = mu ( mg - F sin theta ) `
or ` F ( cos theta + mu sin theta ) = mu mg `
` therefore F = ( mu mg ) /( cos theta + mu sin theta ) `
Work W = Fs ` cos theta `
` therefore W = ( mu mg d cos theta ) /( cos theta + mu sin theta ) `
` ( because s = d ) `