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2 years ago

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A 0.57 kg rubber ball has a speed of 1.2 m/s at point A and kinetic energy 7.5 J at point B. Find a) the ball’s kinetic energy a

t A.(b) What is its speed at point B?(c) What is the total work done on the particle as it moves fromA to B?

1 answer:

worty [1.4K]2 years ago

6 0

Answer:

Explanation:

mass m = .57 kg

speed at A ( v₁ ) = 1.2 m /s

a )

kinetic energy at A

= 1/2 x mass x velocity

= 1/2 x .57 x (1.2)²

= .41 J

b ) Let speed at B be v₂

given that

1/2 m x v₂² = 7.5

.5 x .57 v₂² = 7.5

v₂ = 5.13 m/s

c ) Work done on the particle as it moves from A to B

= increase in kinetic energy

= 7.5 - .41

= 7.09 J

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