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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:
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]
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