Question

problem 7: you throw a ball of mass 0.12 kg up in the air at a speed of 14 m/s. find the initial kinetic energy of the ball upon release. find the work done by gravity on the ball. finally use the work energy theorem to find the height the ball raises up. describe in words what happens to the ball as it raises in the air using the concepts of work and kinetic energy.

Explanation:

Step1: Calculate initial kinetic energy

The formula for kinetic energy is $K = \frac{1}{2}mv^{2}$. Given $m = 0.12\ kg$ and $v=14\ m/s$.
$K=\frac{1}{2}\times0.12\times14^{2}=\frac{1}{2}\times0.12\times196 = 11.76\ J$

Step2: Calculate work done by gravity

At the maximum - height, the final velocity $v_f = 0$. The change in kinetic energy $\Delta K=K_f - K_i$. Since $K_f = 0$, $\Delta K=- 11.76\ J$. According to the work - energy theorem $W=\Delta K$, so the work done by gravity $W=-11.76\ J$.

Step3: Find the height the ball raises

The work done by gravity $W = -mgh$. We know $W=- 11.76\ J$, $m = 0.12\ kg$ and $g = 9.8\ m/s^{2}$.
$-11.76=-0.12\times9.8\times h$. Solving for $h$ gives $h=\frac{11.76}{0.12\times9.8}=10\ m$

Step4: Describe the motion

As the ball rises in the air, the force of gravity acts in the opposite direction of the ball's motion. Gravity does negative work on the ball. This negative work causes a decrease in the ball's kinetic energy. As the kinetic energy decreases, the speed of the ball decreases until it reaches zero at the maximum height.

Answer:

Initial kinetic energy: $11.76\ J$
Work done by gravity: $-11.76\ J$
Height: $10\ m$
Description: Gravity does negative work, decreasing kinetic energy and speed until the ball stops at the maximum height.