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A 5kg block is pushed 3m at a constant velocity up a vertical wall by a constant force applied at an angle of 37 degree with the horizontal, if the coefficient of kinetic friction between the block and the wall is 0.3. what is the work done by the force on the block?

Sagot :

whitneytr12

The work done by the force on the block is:

[tex]W=135.41*3=406.23 \: J[/tex]

The direction of motion here is in the y-direction, so we need to find the component of the force in this direction.

Using trigonometric functions:

[tex]F_{y}=Fsin(37)[/tex]

This component is in the positive y-direction.

Now, there is a friction force between the block and the wall, so we need to find the friction force.

[tex]F_{f}=\mu N[/tex]

Where:

  • μ is the coefficient of kinetic friction
  • N is the normal force (Here N = Fcos(37))

Let's use the first Newton's Law to get F.

F(y) is upward (+) and F(f) and the weight is downward (-).

[tex]F_{y}-F_{f}-W=0[/tex]

[tex]F_{y}-F_{f}=mg[/tex]

Factoring F in the left side:

[tex]F(sin(37)-\mu cos(37))=mg[/tex]    

[tex]F=\frac{mg}{sin(37)-\mu cos(37)}[/tex]  

[tex]F=\frac{5*9.81g}{sin(37)-(0.3*cos(37))}[/tex]  

[tex]F=135.41\: N[/tex]    

Therefore, the work done will be:

[tex]W=Fd[/tex]  

[tex]W=135.41*3=406.23 \: J[/tex]  

You can learn more about work definition here:

https://brainly.com/question/17231715          

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