Answer:
[tex]W_{net}=36J[/tex]
Explanation:
Given data
Mass m=2.0 kg
Magnitude of force F=5.0 N
Initial velocity Vo=8.0i m/s
Final velocity Vf=10.0j m/s
The change in the kinetic energy of canister equals to net work done on the canister
So
ΔK=Wnet
Kf-Ki=Wnet
For initial kinetic energy
[tex]K_{i}=\frac{1}{2}mv_{i}^2\\K_{i}=\frac{1}{2}(2.0kg)(8.0m/s)^2\\K_{i}=64J[/tex]
For final Kinetic energy
[tex]K_{f}=\frac{1}{2}mv_{f}^2\\K_{f}=\frac{1}{2}(2.0kg)(10m/s)^2\\K_{f}=100J[/tex]
Work done on canister by 5.0N force is given as:
[tex]W_{net}=K_{f}-K_{i}\\W_{net}=100J-64J\\W_{net}=36J[/tex]
