Final answer:
The work done on a 3.0 kg canister by a 5.0 N force, when the canister's velocity changes from 5.4 m/s in the positive x-direction to 7.4 m/s in the positive y-direction, is calculated to be 38.52 joules.
Step-by-step explanation:
To determine how much work is done on a canister by a 5.0 N force during its movement in the xy plane, we use the work-energy theorem, which states that the work done by all forces acting on an object is equal to the change in its kinetic energy. The kinetic energy (KE) is given by the equation KE = ½mv², where m is the mass and v is the velocity of the object.
Initially, the canister has a velocity of 5.4 m/s in the positive x-direction, so its initial kinetic energy (KE₁) is ½(3.0 kg)(5.4 m/s)². Later, it has a velocity of 7.4 m/s in the positive y-direction, and its final kinetic energy (KE₂) is ½(3.0 kg)(7.4 m/s)². The work done (W) by the 5.0 N force is the change in kinetic energy, W = KE₂ - KE₁.
Now, we calculate the initial and final kinetic energy and find the difference to get the work done:
- KE₁ = ½(3.0 kg)(5.4 m/s)² = 43.74 J
- KE₂ = ½(3.0 kg)(7.4 m/s)² = 82.26 J
- W = KE₂ - KE₁ = 82.26 J - 43.74 J = 38.52 J
Hence, the work done on the canister by the 5.0 N force is 38.52 J.