A block of mass m takes time t to slide down on a smooth inclined plane of angle of inclination theta and height h. If same block slids down on a rough inclined plane of same angle of inclination and some height and takes time n times of initial value then coefficient of friction b/w block and inclined plane is ????
Well, there is a hard way and an easy way.
Easy way
height = h
angle = A
with no friction
KE gained in slide = m g h
(1/2) m v^2 = m g h
v = sqrt (2 g h)
average speed = v/2 = .5 sqrt (2 g h)
time = average speed * h/sin A
t = .5 h sqrt (2 g h) /sin A
now with work done against friction
Ke = m g h - mu m g cos A (h/sinA)
(1/2) m v^2 = m g h (1 - mu cot A)
v = sqrt [2 g h (1 - mu cot A)]
average speed = v/2
time = .5 h sqrt [2 g h(1 - mu cot A)]/sin A
so
time ratio = n
= sqrt (1 - mu cot A)/sqrt(1)
n^2 = 1 - mu cot A
mu cot A = 1- n^2
mu = (1 - n^2)/cotA
CHECK MY ALGEBRA !!!
Your explanation is good but [1+1/n^2]tanA is the answer . Can you explain how ? I am confused .
To find the coefficient of friction between the block and the rough inclined plane, we can use the concept of energy.
Let's assume that the block initially converts all its potential energy to kinetic energy as it slides down the smooth inclined plane. The potential energy of the block at height h is given by mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.
Now, the kinetic energy of the block can be given as (1/2)mv^2, where v is the velocity of the block at the bottom of the smooth inclined plane.
Since the block slides without friction on the smooth inclined plane, the work done by the force of gravity on the block is equal to the change in kinetic energy. Therefore, we can write:
mgh = (1/2)mv^2
Simplifying, we can cancel out the mass 'm' on both sides:
gh = (1/2)v^2
Now, let's consider the situation where the block slides down the rough inclined plane. Due to the presence of friction, some of the gravitational potential energy will be converted to thermal energy (due to the work done by friction). If it takes n times longer for the block to slide down on the rough inclined plane compared to the smooth inclined plane, we can relate the energy considerations:
n * mgh = (1/2)mv^2 + work done by friction
From the work-energy theorem, work done by friction can be written as:
muk * m * g * h * cos(theta) = (1/2)mv^2 - mgh
Here, uk is the coefficient of friction between the block and the inclined plane, and cos(theta) is the projection of gravity along the inclined plane.
We can simplify the equation further:
uk * g * h * cos(theta) = (1/2)v^2 - gh
Multiplying both sides by 2:
2uk * g * h * cos(theta) = v^2 - 2gh
Rearranging the equation:
v^2 = 2uk * g * h * cos(theta) + 2gh
Substituting the value of v^2 from the equation derived earlier (gh = (1/2)v^2):
gh = 2uk * g * h * cos(theta) + 2gh
gh - 2gh = 2uk * g * h * cos(theta)
Simplifying further:
-gh = 2uk * g * h * cos(theta)
-1 = 2uk * cos(theta)
Finally, rearranging the equation to isolate the coefficient of friction:
uk = -1 / (2 * cos(theta))
Therefore, the coefficient of friction (uk) between the block and the rough inclined plane is:
uk = -1 / (2 * cos(theta))
To find the coefficient of friction between the block and the inclined plane, we'll start by using the given information and some basic physics principles.
Let's assume that the acceleration of the block on the inclined plane is the same in both cases (on the smooth and rough plane). This assumption allows us to compare the time taken for the block to slide down.
On the smooth inclined plane:
Using the equations of motion, we know that the time taken (t) for the block to slide down is given by:
h = (1/2)gt²
Where h is the height of the inclined plane, g is the acceleration due to gravity, and t is the time taken.
On the rough inclined plane:
Given that the block takes n times the initial time (nt), we can write the equation as:
h = (1/2)g(nt)²
Now, let's compare the two equations:
(1/2)gt² = (1/2)g(nt)²
Cancelling the common terms (1/2)g, we get:
t² = (nt)²
Simplifying further:
t = nt
From this equation, we can see that the initial time (t) is equal to the product of n (the factor by which the time is increased) and the new time (nt) on the rough inclined plane.
Therefore, the coefficient of friction (μ) can be determined using the relation:
μ = tan(theta)
Where theta is the angle of inclination of the inclined plane.
So, the coefficient of friction between the block and the rough inclined plane is μ = tan(theta).