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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, dont worry, no need to download diagram attachment to view it.
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, dont 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.
F=mg
This is primarily to find Fn and Fg and the amount of force is exherted by a mass.
Ff=[itex]\mu[/itex]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
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=[itex]\mu[/itex]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
Dont forget to check the diagram, as it will probably help you see the problem correctly, dont 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, dont 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=[itex]\mu[/itex]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=[itex]\mu[/itex]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
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