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2 years ago

8

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

1 answer:

amm18122 years ago

3 0

Answer:

<h2> 16.79J </h2>

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

$Kintial= \frac{1}{2} mv^2\\\\Kintial= \frac{1}{2}*2*2.5^2\\\\KInitial= \frac{1}{2} *2*6.25\\\\Kinitial= 6.25J$

K.E final

$Kfinal= \frac{1}{2} mv^2\\\\ Kfinal= \frac{1}{2}*2*4.8^2\\\\ Kfinal= \frac{1}{2} *2*23.04\\\\ Kfinal= 23.04J$

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

$W net= 23.04-6.25= 16.79J$

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