Question

The only force acting on a 2.0 kg canister that is moving in an xy plane has a magnitude of 5.0 N. The canister initially has a velocity of 8.0 m/s in the positive x direction and some time later has a velocity of 10.0 m/s in the positive y direction. How much work is done on the canister by the 5.0 N force during this time

Answer 1

`W_(net)=36J`

Step-by-step 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

`K_(i)=(1)/(2)mv_(i)^2\\K_(i)=(1)/(2)(2.0kg)(8.0m/s)^2\\K_(i)=64J`

For final Kinetic energy

`K_(f)=(1)/(2)mv_(f)^2\\K_(f)=(1)/(2)(2.0kg)(10m/s)^2\\K_(f)=100J`

Work done on canister by 5.0N force is given as:

`W_(net)=K_(f)-K_(i)\\W_(net)=100J-64J\\W_(net)=36J`