Question

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 at 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?

Answer 1

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