Finding the acceleration of a block on an inclined plane with friction

[Patrica87]

Homework Statement

A block of mass 5 kg is placed on a 30° incline. The coefficient of kinetic friction between the block and the surface is 0.2. Find the acceleration of the block as it slides down the incline.

Relevant Equations

Fnet=maF_{\text{net}} = maFnet=ma

Fgravity, parallel=mgsin⁡θF_{\text{gravity, parallel}} = mg\sin\thetaFgravity, parallel=mgsinθ

Ffriction=μmgcos⁡θF_{\text{friction}} = \mu mg\cos\thetaFfriction=μmgcosθ

Attempt at a Solution:
I started by identifying the forces acting on the block. The component of gravity parallel to the incline is:


Fgravity, parallel=(5)(9.8)sin⁡(30∘)=24.5 NF_{\text{gravity, parallel}} = (5)(9.8)\sin(30^\circ) = 24.5\text{ N}Fgravity, parallel=(5)(9.8)sin(30∘)=24.5 N


The frictional force is:


Ffriction=μmgcos⁡(30∘)=(0.2)(5)(9.8)(0.866)≈8.49 NF_{\text{friction}} = \mu mg\cos(30^\circ) = (0.2)(5)(9.8)(0.866) \approx 8.49\text{ N}Ffriction=μmgcos(30∘)=(0.2)(5)(9.8)(0.866)≈8.49 N


So the net force down the incline should be:


Fnet=24.5−8.49≈16.01 NF_{\text{net}} = 24.5 - 8.49 \approx 16.01\text{ N}Fnet=24.5−8.49≈16.01 N


Then by Newton’s second law:


a=Fnetm=16.015≈3.2 m/s2a = \frac{F_{\text{net}}}{m} = \frac{16.01}{5} \approx 3.2\text{ m/s}^2a=mFnet=516.01≈3.2 m/s2


I’m not certain if this is correct — am I accounting for friction in the right way? Also, should the friction force be included as positive or negative in the net force calculation? Any guidance would be appreciated!

 

[Steve4Physics]

I’m not certain if this is correct

Welcome to PF! I agree with your results but have some comments.

am I accounting for friction in the right way? Also, should the friction force be included as positive or negative in the net force calculation? Any guidance would be appreciated!

Yes, you have accounted for friction correctly.

It is good practice to draw a free body diagram – it will help to identify the forces and get their signs correct.

The frictional force opposes the sliding motion. Here, taking downhill as positive, the frictional force points uphill and is therefore negative.

You might ask yourself what the acceleration would be if the block were moving upwards (e.g. after an initial uphill push).
________________________

Why have you repeated the calculation at each step? E.g.

Fgravity, parallel=(5)(9.8)sin⁡(30∘)=24.5 NF_{\text{gravity, parallel}} = (5)(9.8)\sin(30^\circ) = 24.5\text{ N}Fgravity, parallel=(5)(9.8)sin(30∘)=24.5 N

- you have done the calculation three times (with the same result)!

I would advice you to do the working entirely with symbols (no intermediate calculations). You then end up with a formula for acceleration in terms of $\mu, g, \theta$ and $m$. (You might even find that $m$ cancels-out!) Then, as the final step, put your values in to calculate the acceleration.

You might also want to choose shorter/simpler symbols for the forces. E.g. you could call the component of weight parallel to the slope $F_{par}$ - as long as the meaning is clear.

Minor edits.

 

[kuruman]

In addition to @Steve4Physics's comments:

1. Please note that we prefer that you post your work in $\LaTeX$ which makes the equations more legible

and easier to refer to. To learn how, please click on the link "LaTeX Guide", lower left above "Attach files." To preview (and possibly edit) your equation(s) befpre posting, click the "Preview", button, upper right.

2. I recommend that you put separate equations on separate lines. This

Fgravity, parallel=mgsin⁡θF_{\text{gravity, parallel}} = . . .

makes the reader think that one may cancel Fgravity,parallel on both sides of the equation and end up with
1=mgsin⁡θ !!

Then you would be able to write legibly clear equations like
$~F_{g,\parallel}=mg\sin\!\theta$
$~F_{g,\parallel}=5~(\text{kg})\times 9.8~(\text{m/s}^2)\sin(30^{\circ})=24.5~\text{N}.$