A block of mass m is pulled along a horizontal surface by applying a force at an angle theta with the horizontal. If the block travels with a uniform velocity and has a displacement d and the coefficient of friction is mu, then the work done by the applied force is

A block of mass m is pulled along a horizontal surface by applying a f

1.	A block of mass m is pulled along a horizontal surface by applying a force at an angle theta with the horizontal. If the block travels with a uniform velocity and has a displacement d and the coefficient of friction is mu, then the work done by the applied force is
Answer» `(mu mgd)/(cos theta + mu sin theta)` `(mu m g d cos theta)/(cos theta + mu cos theta)` `(mu m g d sin theta)/(cos theta + mu sin theta)` `(mu m g d cos theta)/(cos theta - mu sin theta)` Solution : Because the block moves with a uniform velocity, the resultant force is zero. Resolving F into horizontal components `F cos theta` and vertical component `F sin theta,` we GET `R + F sin theta = mg orR = mg - F sin theta` Also , `f = mu R = mu(mg - F sin theta)` But `f = F cos theta` `:. F cos theta = mu (mg - F sin theta)` or `F (cos theta + mu sin theta) = mumg` or `F = (mu mg)/(cos theta + mu sin theta theta)` work done, `W = FS cos theta, W= (mu m g d cos theta)/(cos theta + mu sin theta) ( :. s = d)`