Algebra Based Physics Hw Help?

Question

A 4.47 kg block initially at rest is pulled to the right along a horizontal surface by a constant, horizontal force of 15.4 N. The coefficient of kinetic friction is 0.133.
The acceleration of gravity is 9.8 m/s2 .
Find the speed of the block after it has moved 3.39 m.

Answer 1

Let us draw the free-body diagram of the block,

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To calculate normal force using equilibrium equation in y-direction

ΣFy=0:

N-W=0

N-mg=0

N=mg

Here m is the mass and g is the acceleration due to gravity.

Substitute m=4.47 kg and g=9.8 m/s^2 in above equation,

N=4.47x9.8=43.806 N

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Frictional force,Ff=μkN

μk=Coefficient of kinetic friction

Ff=0.133x43.806=5.826 N

Using almbert principal in x-direction

ΣFx=ma

Ff=ma---->a=Ff/m

a=5.826/4.47=1.3034 m/s^2

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Using motion equation

v^2-u^2=2as

v^2-0=2x1.3034x3.39

 v^2=8.837

v=2.973 m/s

speed of the block after it has moved 3.39 m.is 2.973 m/s

 

I hope that helps u

 

 

 

 

Comments

thank u very much