Question

A $6kg$ block is kept on an inclined rough surface as shown in the figure. Find the force $F$ required to
$a)$ keep the block stationary.
$b)$ move the block downwards with constant velocity &
$c)$ move the block upwards with an acceleration of $4m/s^2$.(Take $g=10m/s^2$)

Answer 1

(a) N= F+ mgcos${60}^0$

$\text{mgsin}{60}^0=\mu N=\mu [F+\text{mgcos}{60}^0]$

F$=\frac{\text{mgsin}{60}^0}{{\mu }_s}-\text{mgcos}{60}^0=\left(\frac{6\times 10\times \sqrt{3}}{2\times 0.4}\right)-\left(\frac{6\times 10\times 1}{2}\right)=56.5$

(b)$F_{net}=\text{2gsin}{60}^0-f=2g\left(\frac{\sqrt{3}}{2}\right)-{\mu }_k\text{2gcos}{60}^0=g\sqrt{3}-0.4\times 2g\times \left(\frac{1}{2}\right)=10\times 1.33=13.3N$

(c) As we know Force= mass$\times$acceleration

acceleration = 4 m/$s^2$

Now as the object is inclined at an angle of ${60}^0$

ForceF=$6\times 4\times \text{cos}{60}^0$

=12N

Now in order to move the object from its stationary position the force applied must be greater than 56.5 N and as the object is accelerated by 4 $\text{m}/s^2$

Thus force applied to move the objetc will be=56.5+12=68.5 N