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22 February, 09:02

A 15.0-kilogram mass is moving at 7.50 meters per second on a horizontal, frictionless surface. What is the

total work that must be done on the mass to increase its speed to 11.5 meters per second?

(1) 120. J (3) 570. J

(2) 422 J (4) 992 J

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  1. 22 February, 12:00
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    The answer is 570 J. The kinetic energy has the formula of 1/2mV². The total work in this process W = 1/2m (V2²-V1²) = 1/2 * 15.0 * (11.5²-7.50²) = 570 J.
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