Get the answers you need at Westonci.ca, where our expert community is always ready to help with accurate information. Join our Q&A platform to get precise answers from experts in diverse fields and enhance your understanding. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
The magnitude of the work done by force experience by the object is (2a²b + 3b²)J.
Work done by the force experienced by the object
The magnitude of the work done by force experience by the object is calculated as follows;
W = f.d
where;
- F is the applied force (2xyi + 3yj), where x and y are in meters
- d is the displacement of the object = (a, b)
The work done by the force is determined from the dot product of the force and the displacement of the object.
F = (2xyi + 3yj).(a + b)
W = (2abi + 3bj).(ai + bj)
W = (2a²b + 3b²)J
Thus, the magnitude of the work done by force experience by the object is (2a²b + 3b²)J.
The complete question is below:
The particle moves from the origin to the point with coordinates (a, b) by moving first along the x-axis to (a, 0), then parallel to the y-axis.
How much work does the force do?
Learn more about work done here: https://brainly.com/question/8119756
The total work done for an object moving in the xy-plane is subjected to the force f⃗ =(2xyı^ 3yȷ^)n, where x and y are in m is 3ab N.
What is work done?
Work done is the force applied on a body to move it over a distance. Work done for inclined plane can be given as,
[tex]W=F\times d[/tex]
Here (F) is the magnitude of force and (d) is the distance traveled.
An object moving in the xy-plane is subjected to the force
[tex]\vec f =(2xy\hat i +3y\hat j)[/tex]
Here, x and y are in meter.
The particle moves from the origin to the point with coordinates (a, b) by moving first along the x-axis to (a, 0), then parallel to the y-axis.
- (a) How much work does the force do?
For the first part, the particle moves along x-axis. It moves zero along x-axis. Thus, the force, as y=0.
[tex]W_1=\int\limits^a_0 {2xy\hat i} \, dx =0\\[/tex]
Now, when the object moves along y-axis,
[tex]W_2=\int\limits^a_0 {3y\hat j} \, dx \\W_2=3\int\limits^b_0 {y\hat j} \, dx\\W_2=3y(a-0)\\W_2=3ba\\W_2=3ab[/tex]
Total work done,
[tex]W=0+3ab\\W=3ab\rm\; N[/tex]
Thus, the total work done for an object moving in the xy-plane is subjected to the force f⃗ =(2xyı^ 3yȷ^)n, where x and y are in m is 3ab N.
Learn more about the work done here;
https://brainly.com/question/25573309
#SPJ4
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.