Application of Newtons laws (Friction related) check my solutions please

[supernova1203]

Application of Newtons laws (Friction related) check my solutions please :)

Dont forget to check the diagram, as it will probably help you see the problem correctly, don't worry, no need to download diagram attachment to view it.

Homework Statement

A child pushes a block of wood with a mass of 0.72 kg across a smooth table. The block starts from a position of rest and after 2 seconds its has a velocity of 1.6 m/s [forward]
The coefficient of friction is 0.64.

35a) Draw a free body diagram
- I drew the diagram and its in the attachment, don't worry you needn't downloading it to view it-

35b) Find the net force acting on the block of wood.

35c) Find the force of friction acting on the block of wood.

35d) Find the force with which the child actually pushes on the block of wood.

Homework Equations

F=mg
This is primarily to find Fn and Fg and the amount of force is exherted by a mass.

Ff=$\mu$Fn

This is to find force of friction, but strangely enough a lot of times in the course it can also be used to find force applied...this is something that has always perplexed me

The Attempt at a Solution

35a)
we use F=mg to find force normal, which is also Fg

=(0.72)(9.8)
F=7.056 N


we use the coefficient of friction and Fn to find force of friction, which is also force applied.

Ff=$\mu$Fn

Ff=(0.64)(7.056)

Ff= 4.51 N

Using this information we added magnitudes to our free body diagram (actual numbers in addition to just the directions/vectors) as shown in the diagram.


35b)


Fnet=0 N

Since
F left - F right = 0 N


35c)
Ff=4.51 N as we have already found this.


35d)

Fa=4.51 N


Thanks for checking all this, let me know if its right or not, if its not, please tell me how do to fix the solutions :)

thanks

 

[tiny-tim]

hi supernova1203!

for questions like this, you must start by finding the acceleration

(and your free body diagram is wrong …

ma is certainly not equal to the friction)​

 

[supernova1203]

can i use F=ma in this scenario to find net force? Iv used it before successfully in similar problems to find net force.

Using v1= 0 and v2 over T

I can find acceleration, and since i have mass i can re arrange to find force? Which in this case is net force?#edit

Any way you can show me how to find applied force in this case so i can find net force for 35b)?

 

[Kavik]

The block starts from a position of rest and after 2 seconds its has a velocity of 1.6 m/s [forward]

Sounds like the block is accelerating, so there is some net force in that direction. It looks like you've already noticed this and realized you can find the net force (F=ma will work once you get acceleration, look up kinematic equations if necessary). Once you get that, you already calculated the frictional force so you can use those two to find the applied force.

There's a reason they asked for the net force first! :)

And just to be clear, when using F = ma, with 'a' being the total acceleration, then F is the sum of all forces applied to the block. If you were working with vectors, you'd be looking at both X and Y axes when using F = ma. But, you know that acceleration is zero on the Y axis (sitting on ground), so you can just examine the acceleration on the X axis. Calculating force from the X axis acceleration will of course only yield F (sum of the forces) on the X axis. There could be a million forces acting on the block on the X axis but the sum of those forces would be F. In your case there's only two forces on the X axis (friction/applied) summing to get F.

 

[supernova1203]

neato! so i guess i figured it out! Thanks for the confirmation!

 

[tiny-tim]

neato! so i guess i figured it out! Thanks for the confirmation!

but you haven't found the acceleration …

for questions like this, you must start by finding the acceleration

… (F=ma will work once you get acceleration, look up kinematic equations if necessary). Once you get that, …

 

[supernova1203]

i found solved the problem already, my laptop stopped working on me however, and i couldn't post, i also found acceleration :)