Answer:
Explanation:
Given data
mass of canister= 3.4 kg
force acting on canister= 3 N
initial velocity u= 2.5 m/s
final velocity v= 4.8 m/s
The work done on the canister is the change in kinetic energy on the canister
change in [tex]KE= Kfinal- Kinitial[/tex]
K.E initial
[tex]Kintial= \frac{1}{2} mv^2\\\\Kintial= \frac{1}{2}*2*2.5^2\\\\KInitial= \frac{1}{2} *2*6.25\\\\Kinitial= 6.25J[/tex]
K.E final
[tex]Kfinal= \frac{1}{2} mv^2\\\\ Kfinal= \frac{1}{2}*2*4.8^2\\\\ Kfinal= \frac{1}{2} *2*23.04\\\\ Kfinal= 23.04J[/tex]
The net work done is [tex]KE= Kfinal- Kinitial[/tex]
[tex]W net= 23.04-6.25= 16.79J[/tex]
