Answer to Question #104246 in Mechanics | Relativity for Sridhar

Question

A disc of mass 100 gram slides down from rest on an inclined plane of 30° and comes to rest after travelling a distance of 1m along the horizontal plane.If the coefficient of friction is 0.2 for both inclined and horizontal planes then the work done by the frictional force over the whole journey approximately is (Acceleration due to gravity =10ms^(-2) )

Ans: 0.306J

Answer 1

As per the question,

mass of the disc=100 gm=0.1kg

Angle of inclination $\theta=30^\circ$

Let the final velocity of the on the horizontal plane =v_f

Let the initial velocity of the on the inclined plane =u_i

Let the work due to friction on the inclined plane w₁ and work due to friction on the horizontal plane w₂

and net work done due to friction is W

Horizontal distance traveled by the disc=1m

Coefficient of friction $\mu=0.2$

Acceleration due to gravity $(g)=10m/sec^2$

Let the velocity of the object when it was leaving the incline plane is v,

Let net acceleration on the incline plane be a

$a=g\sin \theta-\mu g\cos\theta=g(\sin\theta-\mu \cos\theta)$

$\Rightarrow a=10(\sin30-0.2 \cos 30)=5(1-0.2\sqrt{3})$

$\Rightarrow a=5(1-0.2\sqrt{3})$ =3.264$m/sec^2$

The de-acceleration due to the friction force

$a_s=\mu g=0.2\times10=2m/sec^2$

now $v_f^2=v^2-2as$

$0=v^2-2\times 2\times 1$

$v=2m/sec$

So, now motion on the incline is happening, let the length of the incline plane is l, so

$v^2=u_i^2+2al$

$2^2=0+2\times3.264\times l$

$\Rightarrow l=\dfrac{4}{6.528}$

$\Rightarrow l=0.612m$

Hence the work done due to the friction force

$W=w_1+w_2$

$\Rightarrow W=f_s . 1+(\mu mg\cos\theta ) l$

$\Rightarrow W=(\mu mg ). 1 +(\mu m g \cos 30) \times 0.612$

$\Rightarrow W=0.2\times 0.1\times 10+0.2\times0.1\times10\times\dfrac{\sqrt{3}}{2}\times 0.612$

$\Rightarrow W=0.2+0.106=0.306J$