Answer to Question #104246 in Mechanics | Relativity for Sridhar

Question #104246
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
1
Expert's answer
2020-03-02T10:14:07-0500

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 =vf

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

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"


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS