The work done by the applied force on the object is (2ab²i + 3b²j) J.
<h3>Magnitude of the force on the object</h3>
The magnitude of the force on the object is calculated as follows;
f = (2xyi + 3yj)
when;
x = a, and y = b
f = (2abi + 3bj)
<h3>Work done by the force</h3>
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: brainly.com/question/8119756
