force F(x) = (−6.0x2 + 5.0x) N acts on a particle. A.)How much work (in J) does the force do on the particle as it moves from  x = 7.0 m to x = 9.0 m?   B.)Picking a convenient reference point of the potential energy to be zero at x = 0, find the potential energy for this force. (Use the following as necessary:

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A: stretch in spring (d) = 9 cm = 0.09 m work done = 19 J

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A:

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A: Given: The stretched length of spring is x=9 cm. The work is W=19 J. The compression in spring is…

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A: (a) Given data The expression for the force is: F = (-4x2 + 5x) N The displacement of the particle…

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A: recall Work-kinetic energy theorem ∑W=∆KEHere ∆KE=0 as v=constant

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A: A force acts on a particle,

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Q: A-C please!!

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A: Given data:   Mass of volleyball (m) = 0.25 kg Height of window (h) = 25 m

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Q: A force F(x) = (3.0/x) N acts on a particle as it moves along the positive x-axis. a) How much work…

A: Solution: as given that F(x) = (3/x) a). work done in moving the particle from x = 2.6 m to x = 5.7…

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A: mass of ball = 0.582Kg H0 =6.72 m Ht = 2.11 m

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A: Mass of skier (m) = 70 kg Initial speed (vo) = 1 msHeight of hill (H) = 17 m velocity at the bottom…

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A: Given Mass of the particle m = 100 g  = 0.1 kg Velocity v = 29 m/s

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A: The frictional force =35N  The constant speed=25m