Question

a block slides down a frictionless 30 degre plane in half the time it takes to slide down on an identical rough plane.Then the coefficient of friction between the block and the incline is

Answer 1

@wolf1728 @AlexandervonHumboldt2 @Mimi_x3 @pooja195

Answer 2

@JoannaBlackwelder

Answer 3

@perl @bibby @Nnesha @iambatman @Loser66 @Destinymasha @Kainui

Answer 4

well where do you think you would start?

Answer 5

I have no idea.

Answer 6

I have an another question would you like to try?@pooja195

Answer 7

@jagr2713

Answer 8

Yes

Answer 9

Can you help me @jagr2713

Answer 10

Sure

Answer 11

Then Please help

Answer 12

whats the question

Answer 13

a block slides down a frictionless 30 degree plane in half the time it takes to slide down on an identical rough plane.Then the coefficient of friction between the block and the incline is

Answer 14

sorry i dont understand the question but let me get someone else to help you.
@iambatman @Destinymasha @Directrix

Answer 15

ok

Answer 16

I would suggest setting up two differential equations of motion. Have one consider an incline plane with no frictional force and the other consider a frictional force with a coefficient of kinetic friction mu.

First consider geometry to figure out what component of gravity will be acting on the block.
We see that the block will slide due to mgsin30. The normal force acting on the block will be mgcos30.

I'm not completely confident in this, so hopefully somebody can look this over. But it seems to make sense to me and be pretty straightforward.

Write two equations of motion for each incline and solve for position equation.

Frictionless:\[ma =mg\sin(30)\]\[a=\frac{g}{2}\]\[v=v_{0}+\frac{g}{2}t\]\[x=x_{0}+v_{0}t+\frac{g}{4}t^{2}\]Assume initial position x0 and initial speed 0 m/s.\[x=x_{0}+\frac{g}{4}t^{2}\]
Rough:\[ma=mg\sin(30)-\mu mg\cos(30)\]\[a=\frac{g}{2}-\frac{\sqrt3\mu g}{2}=\frac{g}{2}(1-\sqrt3 \mu)\]\[v=v_{0}+\frac{g}{2}(1-\sqrt3 \mu)t\]\[x=x_{0}+v_{0}t+\frac{g}{4}(1-\sqrt3 \mu)t^{2}\]Assume initial position x0 and initial speed 0 m/s.\[x=x_{0}+\frac{g}{4}(1-\sqrt3 \mu)t^{2}\]

Now, we have two equations. We can assume that the block reaches the bottom of the incline (x = 0) for both equations. We can also substitute time 't' into the rough incline equation and 't/2' into the frictionless incline equation.

Frictionless:\[0=x_{0}+\frac{g}{4}(\frac{t}{2})^{2}=x_{0}+\frac{g}{16}t^{2}\]
Rough:\[0=x_{0}+\frac{g}{4}(1-\sqrt3 \mu)t^{2}\]

We equate the two equations and solve for coefficient of friction mu:\[x_{0}+\frac{g}{16}t^{2}=x_{0}+\frac{g}{4}(1-\sqrt3 \mu)t^{2}\]\[\frac{g}{16}t^{2}=\frac{g}{4}(1-\sqrt3 \mu)t^{2}\]\[\frac{1}{4}=1-\sqrt3 \mu\]\[\mu =\frac{\sqrt3}{12}\]

So that's what I find for the coefficient. Hope this was the proper approach.