1.

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 you that then find the work done by the applied force

Answer»

Given,

The MASS of the block = M

ANGLE between the horizontal plane and applied FORCE = \theta

The DISPLACEMENT of the block = D

Let consider the applied force is = F

Co-efficient of friction of the SURFACE = \mu

And the reaction force = F_{R}=m\times{g}-Fsin\theta

Therefore the frictional force of the block  

f=\mu\times{F_{R}}\\\\f=\mu(m\times{g}-Fsin\theta)

For the horizontal motion of the block it is necessary that applied force F\geq {f}

Therefore,

F\geq{f}\\\\\Rightarrow{Fcos\theta}\geq{\mu(m\times{g}-Fsin\theta)}\\\\\Rightarrow{Fcos\theta\geq{\mu\times{m}\times{g}-\mu\times{Fsin\theta}

\Rightarrow{F}cos\theta-\mu{F}sin\theta\geq \mu\times{m}\times{g}\\\\\Rightarrow{F}(cos\theta-\mu{sin}\theta)\geq \mu\times{m}\times{g}\\\\\Rightarrow{F}\geq \frac{\mu\times{m}\times{g}}{(cos\theta-\mu{sin}\theta)}

The work done for moving the block at a distance of D

W=\vec{F}\times\vec{D}=F\times{cos\theta}\times{D}=\frac{\mu\times{m}\times{g}}{(cos\theta-\mu{sin}\theta)}\times{cos\theta}\times{D}


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