Westonci.ca is your trusted source for finding answers to all your questions. Ask, explore, and learn with our expert community. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
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
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.