How to find the magnitude of the acceleration given F, kF, m

[Nairb]

<Moved from a technical forum and therefore no template.>

A block is pushed up a 30 incline by an applied force as shown. If F = 50 N and m = 4.41 kg and the coefficient of kinetic friction is 0.4. What is the magnitude of the resulting acceleration on the block?

This is what I did but it's wrong
Fnet= Fa-Fs
m*a= 50-0.2*4.41*9.8*cos(30 and then divide the whole thing by 4.41.
but the answer is wrong. What did I do wrong?

 

[TomHart]

Welcome to Physics Forums.
I apologize, but I am having a difficult time understanding what you did. I will try to explain a couple of things. The force F = 48.5 is not acting in the direction of the acceleration. Only a component of it is. Also, the F = 48.5 N force also has the effect of increasing the normal force because a component of force F is acting perpendicular (edit: downward) to the plane of the surface.

Did you draw a free body diagram? For these types of problems, it is extremely helpful to draw a free body diagram. For this problem, I would assign the x-axis along the slope of the plane, and the y-axis perpendicular to that. That way, your acceleration will be entirely along the x axis.

Once you have your free body diagram, you will need to sum forces in the x direction and sum forces in the y direction. Then you can solve those 2 equations for the acceleration.

 

[gneill]

@Nairb, Is there an image to accompany the problem? It hard to check what's been tried without having a clear understanding of the actual scenario.

 

[TomHart]

Is there an image to accompany the problem?

Hmmm. There was an image originally. It seems that when the values were edited, the image disappeared. The force F was acting horizontally on the block that was sitting on a 30° incline.

 

[Nairb]

Sorry, the image was deleted by accident. Here it is:

what I did was this: F - (uk*m*g*cos30) divided by the mass

 

[haruspex]

F - (uk*m*g*cos30)

You need to analyse the balance of forces on the block more carefully.
What forces have components in the direction normal to the slope? What equation does that give?
Likewise for parallel to the slope.