# Thread: Calculate the velocity of the brick using an assumption & draw the displacement-time graph

1. Calculate the velocity of the brick at the bottom of the inclined plane.

& Sketch the displacement-time graph relevant to the movement of the brick along the smooth gutter. (Assume that the brick started to move from the state of rest.)

Workings:

I am unable to think on the first two but I think the displacement and time graph should be something like ,

Many Thanks

2. To calculate the velocity of the brick at the bottom of the gutter, I would take the statement that the gutter is smooth to mean there is no friction, and so conservation of energy can be applied. When the brick is at the top of the gutter, it has gravitational potential energy, and since it begins from rest it has no kinetic energy. Then when the brick is at the bottom of the gutter it has kinetic energy, but no gravitational energy. The amount of energy stays the same, but it changes from potential energy to kinetic energy.

So, if we equate the initial potential energy to the final kinetic energy, we have:

$\displaystyle mgh=\frac{1}{2}mv^2$

What do you get upon solving for $v$?

Originally Posted by MarkFL
To calculate the velocity of the brick at the bottom of the gutter, I would take the statement that the gutter is smooth to mean there is no friction, and so conservation of energy can be applied. When the brick is at the top of the gutter, it has gravitational potential energy, and since it begins from rest it has no kinetic energy. Then when the brick is at the bottom of the gutter it has kinetic energy, but no gravitational energy. The amount of energy stays the same, but it changes from potential energy to kinetic energy.

So, if we equate the initial potential energy to the final kinetic energy, we have:

$\displaystyle mgh=\frac{1}{2}mv^2$

What do you get upon solving for $v$?
Thank you very much MarkFL

$\displaystyle mgh=\frac{1}{2}mv^2$

$\displaystyle 2kg * 10 ms^-2 * 5 m=\frac{1}{2}mv^2$

$\displaystyle 2kg * 10 ms^-2 * 5 m=\frac{1}{2}*2kg * v^2$

$\displaystyle 2kg * 10 ms^-2 * 5 m=\frac{1}{2}*2kg * v^2$

$\displaystyle 100 J= v^2$

$\displaystyle 10 J= v$

Correct ?

4. I would solve the equation first, and then plug in the numbers:

$\displaystyle mgh=\frac{1}{2}mv^2$

$\displaystyle v=\sqrt{2gh}$

Okay, at this point we can plug in:

$\displaystyle g\approx9.81\,\frac{\text{m}}{\text{s}^2},\,h=5\text{ m}$

$\displaystyle v\approx\sqrt{2\left(9.81\,\frac{\text{m}}{\text{s}^2}\right)\left(5\text{ m}\right)}=3\sqrt{\frac{109}{10}}\,\frac{\text{m}}{\text{s}}\approx9.9045\,\frac{\text{m}}{\text{s}}$

Originally Posted by MarkFL
I would solve the equation first, and then plug in the numbers:

$\displaystyle mgh=\frac{1}{2}mv^2$

$\displaystyle v=\sqrt{2gh}$

Okay, at this point we can plug in:

$\displaystyle g\approx9.81\,\frac{\text{m}}{\text{s}^2},\,h=5\text{ m}$

$\displaystyle v\approx\sqrt{2\left(9.81\,\frac{\text{m}}{\text{s}^2}\right)\left(5\text{ m}\right)}=3\sqrt{\frac{109}{10}}\,\frac{\text{m}}{\text{s}}\approx9.9045\,\frac{\text{m}}{\text{s}}$
Thank you very much

Originally Posted by mathlearn

Sketch the displacement-time graph relevant to the movement of the brick along the smooth gutter. (Assume that the brick started to move from the state of rest.)

Is the above drawn graph correct?

6. For the graph, you have the brick returning to its original position, since the ending value is zero. Also, the acceleration will be constant as it moves down the gutter, so what kind of curve should we use?

7. Originally Posted by mathlearn

Is the above drawn graph correct?
Hey mathlearn!

Here's what the acceleration and speed graphs would look like (using our new TikZ drawing capabilities ):

On the inclined plane we have a constant acceleration, and when it ends, the acceleration becomes near zero.
As a result the speed increases linearly until the end of the slope, after which the speed remains constant (although in reality it would linearly decrease slowly until zero).
Note that $v=\int_0^t a\,dt$.

However, we're asked for the displacement, which we can call $d$.
We have that $d=\int_0^t v\,dt$.
What would the displacement graph look like?

8. Originally Posted by mathlearn
$\displaystyle 100 J= v^2$

$\displaystyle 10 J= v$
Recheck your units, speed is not measured in J. (And the square root of a J is not J!)

-Dan

Originally Posted by MarkFL
For the graph, you have the brick returning to its original position, since the ending value is zero. Also, the acceleration will be constant as it moves down the gutter, so what kind of curve should we use?

This kind of a curve with constant acceleration and with deceleration.

Originally Posted by I like Serena
Hey mathlearn!

Here's what the acceleration and speed graphs would look like (using our new TikZ drawing capabilities ):

On the inclined plane we have a constant acceleration, and when it ends, the acceleration becomes near zero.
As a result the speed increases linearly until the end of the slope, after which the speed remains constant (although in reality it would linearly decrease slowly until zero).
Note that $v=\int_0^t a\,dt$.

However, we're asked for the displacement, which we can call $d$.
We have that $d=\int_0^t v\,dt$.
What would the displacement graph look like?
The displacement time graph would look like the above graph

Originally Posted by topsquark
Recheck your units, speed is not measured in J. (And the square root of a J is not J!)

-Dan
Thanks for the catch

10. Originally Posted by mathlearn
This kind of a curve with constant acceleration and with deceleration...
Why do you have the brick returning to zero displacement?