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Sagot :
The work done on the object moving in the xy-plane that is subjected to the given applied force is determined as (2ab²i + 3b²j) J.
Magnitude of the force on the object
The magnitude of the force on the object is calculated as follows;
f = (2xyi + 3yj)
when;
x = a, and y = b
f = (2abi + 3bj)
Work done by the force
The work done the applied force is the product of force and displacement of the object.
W = fΔs
where;
- Δs is displacement of the object
Δx = a - a = 0
Δy = 0 - b = -b
Δs = √(Δx² + Δy²)
Δs = √(-b)²
Δs = b
W = (2abi + 3bj) x b
W = (2ab²i + 3b²j) J
Thus, the work done by the applied force on the object is (2ab²i + 3b²j) J.
The complete question is below;
An object moving in the xy-plane is subjected to the force f = (2xyi + 3yj), where x and y are in m. 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 work done on an item moving in the xy-plane that is subjected to a particular force is found as (2ab²i + 3b²j) J.
What is work done?
Work done is defined as the product of applied force and the distance through which the body is displaced on which the force is applied.
Work may be zero, positive, and negative.it depends on the direction of the body displaced. if the body is displaced in the same direction of the force, it will be positive.
The given data in the problem is;
F is the force applied; f⃗ = (2xyi + 3yj)
Work done is found as the product of force and the body displacement;
[tex]\rm W = F \triangle S[/tex]
As the object is moving in the xy- plane. Displacement in the x direction;
[tex]\rm \triangle x = a-a \\\\ \triangle x =0[/tex]
Displacement in the y direction;
[tex]\rm \triangle y = 0-b \\\\ \rm \triangle y = -b \\\\[/tex]
The net displacement in the xy plane is;
[tex]\rm \triangle s= \sqrt{(\triangle x)^2+(\triangle y)^2 } \\\\\ \rm \triangle s= \sqrt{(0)^2+(-b)^2 } \\\\\ \triangle s= \sqrt{b^2} \\\\ \triangle s=b[/tex]
The work done is found as;
[tex]\rm W= F \triangle S \\\\ W=( 2 ab \vec i+3b \vec j) \triangle S\\\\ W= W=( 2 ab \vec i+3b \vec j) b \\\\\ W=(2ab^2 \vec i + 3b^2 \vec j)\ J[/tex]
Hence, the work done on an item moving in the xy-plane will be,[tex]\rm W=(2ab^2 \vec i + 3b^2 \vec j)\ J[/tex]
To learn more about the work done, refer to the link ;
https://brainly.com/question/3902440
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