Westonci.ca is your trusted source for finding answers to a wide range of questions, backed by a knowledgeable community. Connect with professionals ready to provide precise answers to your questions on our comprehensive Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
The work done by the two forces in moving the chandelier to the given distance is 5,091.2 J.
Work done by a single force on the chandelier
The work done by a single force on the chandelier is determined by taking the net horizontal force applied on the chandelier over the given distance.
[tex]W = (F - F_f) d\\\\W = (F sin\theta \ - \ F_f)d[/tex]
where;
[tex]F_f[/tex] is frictional force = 0 (smooth surface).
[tex]W = Fsin(\theta) d\\\\W = 1200 \times sin(45) \times 12\\\\W = 10,182.34 \ J[/tex]
Work done by the two forces
When the two forces combine to pull the chandelier, the total work done will be shared by the two forces.
[tex]W_1 = W_2 = \frac{10,182.34}{2} = 5,091.2 \ J[/tex]
Thus, the work done by the two forces in moving the chandelier to the given distance is 5,091.2 J.
Learn more about work done here: https://brainly.com/question/8119756
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.