Question

a mass 10 kg slides down an inclined plane which is 1m long and 0.5m high. force of friction is 5N. with what speed it reaches the bottom?​

Answer 1

your answers is in point form 20.2 meter per second

Answer 2

Given:

m = 10kg

h = 0.5m

l = 1m

Force of friction = F = 5N

To find:

Final speed=v

Solution:

Consider the image attached and different forces acting on the body.

$sin\theta = h/l = 0.5/1$

$\Rightarrow$ $\theta = 30 deg$

$F_i = mgsin\theta$  ( force due to inertia)

    $= 10\times 9.8\times 0.5$

    $= 49 N$

Net force  , $F = F_i -Ff$

                       $= 49-5 = 44N$

Now, $F=ma$

         $a= 44 / 10$ $=4.4 m/s^2$

Consider initial velocity , $u = 0$

substituting in the equation ; $v^2 = u^2 +2al$

                                                $v^2 = 2 \times4.4 \times 1$ $=8.8$

                                                $v = 2.966 m/s$

Therefore speed when it reaches the bottom $= 2.966m/s$