# How do I find the horizontal distance of a block when it is sliding down an incline?

#### Chemist116

The problem is as follows:

A body whose mass is $1\,kg$ is sliding over an incline of $5\,m$ in length. One of its sides has an elevation of $3\,m$ with respect of the floor. The other is resting in the floor. Find the horizontal distance which the body will travel before it stops. The coefficient of friction is constant all the path and is $0.125$.

The alternatives given in my book are:

$\begin{array}{ll} 1.&20\,m\\ 2.&2.5\,m\\ 3.&40\,m\\ 4.&28\,m\\ 5.&24.5\,m\\ \end{array}$

This problem doesn't offer a drawing or any sketch. So I made one by my interpretation of the words stated.

As it can be seen it was pretty simple to spot that they were referring to a $3-4-5$ right triangle. However the part where it gets tricky is. How to find the acceleration or something like this in order to find the horizontal distance?.

My initial idea was to use this formula:

$v_{f}^2=v_{o}^2-2a\Delta x$

Since it mentions that it will stop I can assume that the final speed is $0$. But in this case There isn't given a initial speed or anything that I could use.

What would be the answer?. What should be done to solve this problem?. Can somebody help me here?. :help:

#### DarnItJimImAnEngineer

Don't forget potential energy changes. I agree, though, I'm not sure how to solve the problem without making assumptions about the initial velocity.

#### romsek

Math Team
what book is this?

The problem specifies a 5m length of the surface the block is moving on yet 4 of the 5 answers are lengths greater than this.

Without an initial velocity this problem can't be solved.

#### DarnItJimImAnEngineer

Plugging in numbers, the projected weight (along the inclined surface) is greater than the force from sliding friction, so as long as it is moving, it will always accelerate to the end. I assume it then slides along the floor until friction dissipates the kinetic energy.

If we assume the initial kinetic energy to be negligible (or assume it is zero, and that the coefficient of static friction is small enough that it starts sliding on its own), then
$E_i - W_{friction,incline} - W_{friction,floor} = E_f$
$\rightarrow mgh - \mu mg\sin(37Â°)(5~m) - \mu mgL_{floor} = 0$
$\displaystyle \rightarrow L_{floor} = \frac{h}{\mu} + (5~m)\sin(37Â°) = \frac{3~m}{0.125}+(5~m)\sin(37Â°) = 27.0~m$
Add the $4~m$ it slid on the incline, and it has travelled 31.0 m in the horizontal direction.

Unfortunately, this is not one of the answers, so unless I made a mistake (entirely likely at 1 in the morning), or the author made a mistake (also entirely possible based on some of the problems we've seen so far), I can only assume this is not what they intended.

1 person

#### Chemist116

Plugging in numbers, the projected weight (along the inclined surface) is greater than the force from sliding friction, so as long as it is moving, it will always accelerate to the end. I assume it then slides along the floor until friction dissipates the kinetic energy.

If we assume the initial kinetic energy to be negligible (or assume it is zero, and that the coefficient of static friction is small enough that it starts sliding on its own), then
$E_i - W_{friction,incline} - W_{friction,floor} = E_f$
$\rightarrow mgh - \mu mg\sin(37Â°)(5~m) - \mu mgL_{floor} = 0$
$\displaystyle \rightarrow L_{floor} = \frac{h}{\mu} + (5~m)\sin(37Â°) = \frac{3~m}{0.125}+(5~m)\sin(37Â°) = 27.0~m$
Add the $4~m$ it slid on the incline, and it has travelled 31.0 m in the horizontal direction.

Unfortunately, this is not one of the answers, so unless I made a mistake (entirely likely at 1 in the morning), or the author made a mistake (also entirely possible based on some of the problems we've seen so far), I can only assume this is not what they intended.
I'm still baffed by this question. I'm sure that what you intended to do was right. Still the thing about the initial speed is what I don't get it. Perhaps can you retry a different approach?. :help:

#### skeeter

Math Team
initial potential energy - work done by friction on the incline - work done by friction on the floor = 0

$mgh - \mu \cdot mg\cos{\theta} \cdot (5) - \mu \cdot mg \Delta x = 0$

$h - \mu \cdot \cos{\theta} \cdot (5) = \mu \cdot \Delta x$

$\dfrac{h - \mu \cdot \cos(37) \cdot 5}{\mu} = \Delta x$

$\Delta x = 20 \text{ m}$

1 person