# Inclined planes with friction and no friction

###### tags: `Physics`

[TOc]

We did an example using the little mass that we have in class and a table (65cm wide and we lifted it 20 cm )

## Solved problem

We have the solved problem 2 in page 214

A 5kg body slides down a plane inclined at 30ª to the horizontal. Assuming that it was initially at rest, calculate its speed when it has travelled 10 m in 2 cases

a) if there is no friction

b) if the coefficient of friction between the body and the plane is 0.3

### Calculations in common with both cases

We need to use $h$ in both cases because we need to work with the Potencial energy ($E_p$) so we use _trigonometry_ to handle that we use SOH-CAH-TOA. In this case :

$$ sin {\ 30º} = {O \over H} = {h \over 10}$$
hence

$$ h = 10 · sin {\ 30º} = 5 m $$

Now that we have that we can continue

### No friction

If there is no friction then Mechanical energy is constant: $E_{m1} = E_{m2}$

Those moments (1 and 2) are at the start of the slide and in the end.

And now we remember that mechanical energy($E_m$) is kinetic($E_k$) + potential energy($E_p$) so:

$$ E_{k1} + E_{p1} = E_{k2} + E_{p2}$$

Now we know that the initial velocity is 0, so the initial kinetic energy is 0 because $E_{k1} = {1 \over 2}mv^2$ and if $v =0$ then $E_{k1} = {1 \over 2}m0^2 = 0$

We also know that the final potential energy is 0 because $h$ is $0$

#### What we have left

Once we have taken out the 0 parts of that equation we have this:

$$ E_{p1} = E_{k2} $$

And if we unfold it:

$$ m · g·h_1 = {1 \over 2}m ·v_2^2$$

if we make v the subject we have this formula (see that mass is simplified here):

$$ v_2 = \sqrt{2 ·g·h}$$

And if we put our data here...

$$ v_2 = \sqrt{2 · 9.8 m/s^2 · 5m} = 9.9 m/s$$

### Friction

If there is friction, the energy balance at point 1 is diverged into mecanichal energy in the point 2 and work made by the force of friction. In other words:

$$ E_{m1} = E_{m2} + W_{friction}$$

That is:

$$ E_{m1} = E_{m2} + F_{friction} · \Delta x$$

Now we have to find the $F_{friction}$ and to find it we need to find the Normal. The normal is perpendicular to the inclined plane and it's equal to the component of the weight that is also perpendicular to the plane ($W_y$)

$$ F_{friction} = \mu · N = \mu · W_y = \mu · W cos \ 30º$$

And, of course, Weight is mass times g so
$$ F_{friction} = \mu · m · g · cos \ 30º $$

Then we develop the other parts of the Mechanical Energy

$$ E_{m1} = E_{k1} + E_{p1} = 0 + m · g·h_1$$

$$ E_{m2} = E_{k2} + E_{p2} = {1 \over 2}m ·v_2^2 + 0$$

And then we substitute in

$$ E_{m1} = E_{m2} + F_{friction} · \Delta x$$

Hence:

$$ m · g·h_1 = {1 \over 2}m ·v_2^2 + \mu · m · g · cos \ 30º · \Delta x$$

We neet to isolate here $v_2^2$ so we make it the subject:

$$ v_2 = \sqrt{2 ·g·h - \mu · g · cos \ 30º · \Delta x}$$

Note: We can substitute or we can take the mass out of this equation (as we did in the previous example without friction).

## Exercise

We are throwing a ball (2kg) through hill that has a slope of 45º (take it as if it's infinite)

We are throwing it at an initial speed ($v_0$) (in the direction of the slope) of 20 m/s.

Questions:

a) What height ($h_a$) is it going to reach the ball assuming no friction?

b) What is the height($h_b$) is it going to reach the ball assuming a friction of 0.3?

Extra questions:

c) Assuming no friction, what is the speed ($v_{2a}$) that we need to send it to a twice the height $h_a$?

d) what is the relationship between the speed $v_{2a}$ and $v_0$? In other words. How much is $v_{2a} /v_0$?

e) Assuming friction, what is the speed ($v_{2b}$) that we need to send it to a twice the height $h_b$?

f) what is the relationship between the speed $v_{2b}$ and $v_0$? In other words. How much is $v_{2b} /v_0$?

Let's try to dig this out a bit.

### General considerations

In this case we have to consider 2 moments in time that we're considering

- Moment 0 is when we're throwing the ball. Potential Energy is 0 because we are in the bottom part of the slope.

- Moment 1 is when the ball speed is 0, because the ball is going to slow down an then fall again.

This is a moment when it's useful to draw a little scheme if needed.

We also have to be careful. In this case they are asking for the height, but they can ask for how much slope it's going to travel.

### No friction

We know that if there is no friction or external forces, Mechanical Energy is conserved, so $E_{m0}=E_{m1}$.

Mechanical energy in 0 it's only kinetic energy.

Mechanical energy in 1 it's only potential energy.

So we can, in this case make both parts equal.

$$ E_{p1} = E_{k0} $$

And if we unfold it:

$$ m · g·h_a = {1 \over 2}m ·v_0^2$$

In this case we don't have to find $v$, we need to find $h_a$ and we have the rest of the data. So you should find how much is $h_a$.

### Friction

If there is friction, the energy balance at point 0 is diverged into mecanichal energy in the point 1and work made by the force of friction. In other words:

$$ E_{m0} = E_{m1} + W_{friction}$$

Mechanical energy in moment 0 it's still only kinetic energy, and mechanical energy in moment 1 it's still only potential energy so we can unfold the expression like this:

$$ {1 \over 2}m ·v_0^2= m · g·h_b + F_{friction} · \Delta x$$

Notice that here $\Delta x$ it's the diagonal slope. It's the length of the slope where is the ball going. **Try to make a diagram of this**.

If we use SOH-CAH-TOA you can express $\Delta x$ in terms of $h_b$. Why do we do it? Because if we're looking for it, it's better to express everything in terms of that so once we find it we can find everything else.

In this case, if the slope is 45º, we know that

$$ sin \ 45 = {O \over H} = {h_b \over \Delta x}$$

So $\Delta x= {h_b \over sin \ 45}$ and we can put it in the expression that we have before:

$$ {1 \over 2}m ·v_0^2= m · g·h_b + F_{friction} · {h_b \over sin \ 45}$$

We take common factor of $h_b$:

$$ {1 \over 2}m ·v_0^2= h_b( m · g + · {F_{friction} \over sin \ 45}) $$

Now we can find $F_{friction}$ using the approximation of the Normal.

$$ F_{friction} = \mu · N = \mu · W_y = \mu · W cos \ 45º = \mu · m · g \cos \ 45º $$

And... we put that in our expression

$$ {1 \over 2}m ·v_0^2= h_b( m · g + · {\mu · m · g \cos \ 45º \over sin \ 45}) $$

Here we can just operate (we know all the data but $h_b$) or we can take out the m and also find out that cos 45/sin 45 it's equal to tan 45. But it's not necessary.

**Find $h_b$**

### Extra questions

For the extra questions the good news is that we have all set. For the frictionless we had this:

$$ m · g·h_a = {1 \over 2}m ·v_0^2$$

Now instead of having the velocity and find out the height we do otherwise. And, of course, now it's $2·h_a$, so...

$$ m · g·2·h_a = {1 \over 2}m ·v_{2a}^2$$

For the one that has friction there are more data but it's the same:

$$ {1 \over 2}m ·v_{2b}^2= 2· h_b( m · g + · {\mu · m · g \cos \ 45º \over sin \ 45})$$