Question

The only force acting on a 3.4 kg canister that is moving in an xy plane has a magnitude of 3.0 N. The canister initially has a velocity of 2.5 m/s in the positive x direction, and some time later has a velocity of 4.8 m/s in the positive y direction. How much work is done on the canister by the 3.0 N force during this time

Answer 1

16.79J

Step-by-step 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
`KE= Kfinal- Kinitial`

K.E initial

`Kintial= (1)/(2) mv^2\\\\Kintial= (1)/(2)*2*2.5^2\\\\KInitial= (1)/(2) *2*6.25\\\\Kinitial= 6.25J`

K.E final

`Kfinal= (1)/(2) mv^2\\\\ Kfinal= (1)/(2)*2*4.8^2\\\\ Kfinal= (1)/(2) *2*23.04\\\\ Kfinal= 23.04J`

The net work done is
`KE= Kfinal- Kinitial`

`W net= 23.04-6.25= 16.79J`