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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)
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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

User Augusto Carmo
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2 Answers

5 votes

Answer:570j

Step-by-step explanation:

DeltaKE= W

1/2m(KEf-KEi)

User SequenceGeek
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8.6k points
3 votes
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.
User Scottkosty
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