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A 0.60 kg rubber ball has a speed of 1.5 m/s at point A and kinetic energy 7.6 J at point B Find

a) the ball's kinetic energy at A. Answer in units of J.
b) the ball’s speed at B. Answer in units of m/s.
c)the total work done on the ball as it moves from A to B. Answer in units of J.

User Coisox
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1 Answer

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Final answer:

a) The ball's kinetic energy at point A is 0.675 J. b) The ball's speed at point B is approximately 5.03 m/s. c) The total work done on the ball as it moves from A to B is 6.925 J.

Step-by-step explanation:

a) The kinetic energy of the ball at point A can be calculated using the formula:

Kinetic Energy (K) = 0.5 × mass (m) × velocity (v)²

Given that the mass (m) = 0.60 kg and the velocity (v) = 1.5 m/s, we can substitute these values into the formula to get:

K = 0.5 × 0.60 kg × (1.5 m/s)² = 0.675 J

Therefore, the ball's kinetic energy at point A is 0.675 J.

b) The kinetic energy of the ball at point B is given as 7.6 J. To find the velocity (v) at point B, we need to rearrange the formula for kinetic energy:

Velocity (v) = √(2K/m)

Substituting the values of kinetic energy (K = 7.6 J) and mass (m = 0.60 kg) into the formula:

v = √(2 × 7.6 J / 0.60 kg) = √25.333... m/s ≈ 5.03 m/s

Therefore, the ball's speed at point B is approximately 5.03 m/s.

c) The total work done on the ball as it moves from point A to point B can be calculated using the work-energy theorem:

Work (W) = Change in kinetic energy (ΔK) = KB - KA

Given that KB = 7.6 J and KA = 0.675 J, we can subtract these values to find:

W = 7.6 J - 0.675 J = 6.925 J

Therefore, the total work done on the ball as it moves from A to B is 6.925 J.

User Akash Kinwad
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