Question

You throw a ball of mass 160gupward (Figure 6.79). When the ball is 2mabove the ground, headed upward (the initial state), its speed is 19m/s. Later, when the ball is again 2mabove the ground, this time headed downward (the final state), its speed is 19m/s. What is the change in the kinetic energy of the ball from initial to final state?

Short answer

The change in the kinetic energy of the ball from the initial to the final state is.

Step-by-step solution

Identification of the given data

The given data is listed as follows,

• The mass of the ball is,$m=160\mathrm{g}$
• The initial distance of the ball above the ground is,2m
• The speed of the ball initially is,$v_1=19\mathrm{m}/\mathrm{s}$
• The final distance of the ball above the ground is,2m
• The speed of the ball finally is,$v_f=19\mathrm{m}/\mathrm{s}$

Significance of the kinetic energy

The kinetic energy is described as half of the product of the mass and the square of the velocity. It can be expressed as follows,

$\mathbf{KE}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mv}}^{\mathbf{2}}$

Here,$m$ is the mass of the object and$v$ is the velocity of the object.

Determination of the change in the kinetic energy

The equation of the change in the kinetic energy is expressed as:

$$\begin{gathered} ∆KE=K{E}_1+K{E}_i \\ =\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2 \end{gathered}$$

Here,$m$is the mass of the ball,$v_f$is the final velocity of the ball and$v_i$is the initial velocity of the ball.

Substitute all the values in the above equation.

$$\begin{gathered} ∆KE=\frac{1}{2}\left(160g\right){\left(19\mathrm{m}/\mathrm{s}\right)}^2-\frac{1}{2}\left(160g\right){\left(19\mathrm{m}/\mathrm{s}\right)}^2 \\ =0 \end{gathered}$$

Thus, the change in the kinetic energy of the ball from the initial to the final state is 0.