Question

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 1

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

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

Answer 2

"The magnitude of the vector is given as 7, and the vector perpendicular to the given three vector as A = 2 i + 3 j + 6 k and B = i + j – k. Therefore the perpendicular vector will be $C = A \times B$.

$\Rightarrow C = (2 i + 3 j + 6 k ) \times (i + jk )$

$\Rightarrow C = - 3 i + 8 j + 5 k$.

The magnitude of C will then be  therefore the unit vector of 7 be considered along the vector C as U = vector C / magnitude of C

$=> U = - 3 i + 8 j + 5 k / \sqrt 98$

And also C = 7 U thereby,

$7 U = 7 ( 3 i + 8 j + 5 k / \sqrt 98 )$

$=> 7 U = - 21 i + 56 j + 35 k /\sqrt 98$

$=> 7 U = ( -3 / \sqrt 2 ) i + ( 8 / \sqrt 2 ) j + ( 5 / \sqrt 2 ) k$

=> Therefore the required vector $= - 3 i + 8 j + 5 k / \sqrt 2$"