Question

A 0.57 kg rubber ball has a speed of 2.2 m/sat point A and kinetic energy 8.0 J at pointB. Finda) the ball’s kinetic energy at A.Answer in units of J.015 (part 2 of 3) 10.0 pointsb) the ball’s speed at B.Answer in units of m/s.016 (part 3 of 3) 10.0 pointsc) the total work done on the ball as it movesfrom A to B.Answer in units of J.

Answer 1

Given that the mass of the ball is m = 0.57 kg.

The speed of the ball is

`v_A=2.2\text{ m/s}`

A) The kinetic energy at point A will be

`\begin{gathered} K\mathrm{}E._A\text{ =}(1)/(2)m(v_A)^2 \\ =(1)/(2)*0.57*(2.2)^2 \\ =1.38\text{ J} \end{gathered}`

B) Given that the kinetic energy,

`K\mathrm{}E._B=\text{ 8 J}`

The ball's speed at point B will be

`\begin{gathered} K\mathrm{}E._B=(1)/(2)m(v_B)^2 \\ v_B=\sqrt[]{(2K.E._B)/(m)} \\ =\sqrt[]{(2*8)/(0.57)} \\ =5.298\text{ m/s} \end{gathered}`

C) The total work done on the ball to move from point A to B is

`\begin{gathered} W=K\mathrm{}E._B-K.E._A \\ =8-1.38\text{ } \\ =6.62\text{ J} \end{gathered}`