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vfiekz [6]
3 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
Physics
1 answer:
amm18123 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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My goal is to make sure you understand the problem, which is why I won't be giving you the answer. It'll be more work now, but less work in the future! :)

Hope this helped!

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DISCLAIMER: I am not a professional tutor or have any professional background in your subject. Please do not copy my work down, as that will only make things harder for you in the long run. Take the time to really understand this, and it'll make future problems easier. I am human, and may make mistakes, despite my best efforts. Again, I possess no professional background in your subject, so anything you do with my help will be your responsibility. Thank you for reading this, and have a wonderful day/night!
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