# Question d4e0c

Apr 1, 2016

option (4) i.e$. \frac{\mu E \cos \theta}{\sin \theta + \mu \cos \theta}$

#### Explanation:

If $h$ is the height the object moves up the plane and $l$ is the slant distance moved along the plane and $d$ is the run.
$\frac{h}{l} = \sin \theta$ Force of friction ${F}_{f} = \mu N = \mu m g \cos \theta$
Where $\mu$ is the coefficient of dynamic friction
$m$ is the mass
$g$ is gravity
$\theta$ is the angle above horizontal

Work done against friction:

$\text{work done" = "force due to friction" xx "distance moved } l$
$= \left(\mu \times m g \cos \theta\right) \times l$

When the object comes to stop, part of its Kinetic energy $E$ got converted into its potential energy and rest of it has been used to overcome kinetic friction.

$m g h = E - \mu m g \cos \theta \times l$
$m g l \sin \theta = E - \mu m g \cos \theta \times l$, solving for $l$
$\implies l = \frac{E}{m g \sin \theta + \mu m g \cos \theta}$
Again
$\text{work done" = "force due to friction" xx "distance moved } l$
$= \left(\mu \times m g \cos \theta\right) \times l$
=(muEmgcostheta)/(mgsintheta+mumgcostheta)= (muEcostheta)/(sintheta+mucostheta)#