A 20.0 kg object is at rest, and then speeds up to 16.0 m/s. How much work was done on this system?

asked Jun 24, 2019 139k views

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Work=Change in Kinetic Energy
1/2mVf^2-1/2Vi^2=W
1/2(20)(16)^2-1/2(20)(0)^2=W
2560-0=W
W=2560J

Brian Ensink answered Jun 29, 2019

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