A 28 kg ball initially at rest rolls down a 159 m hill. When the ball reaches the bottom it is traveling at 31.3 m/s.

How much energy is dissipated by friction on the ball?

PEmax=mg*hmax = 28*9.8 * 159=43,630 J.

= KEmax @ bottom of hill with no friction.

KE = 14*31.3^2 = 13,716 J.

Energy lost = 43,630 - 13,716=29,914 J.

497.67

To calculate the amount of energy dissipated by friction on the ball, we need to consider the change in energy from the top to the bottom of the hill.

First, let's determine the potential energy of the ball at the top of the hill. The potential energy (PE) is given by the equation: PE = mass * gravity * height.

Since the ball is initially at rest, it doesn't have any kinetic energy, so all of its energy is potential energy at the top.

PE = 28 kg * 9.8 m/s² * 159 m = 44,319.2 J

Next, let's determine the kinetic energy (KE) of the ball at the bottom of the hill. The kinetic energy is given by the equation: KE = (1/2) * mass * velocity².

KE = (1/2) * 28 kg * (31.3 m/s)² = 14,058.58 J

To find the energy dissipated by friction, we need to find the difference between the initial potential energy and the final kinetic energy:

Energy dissipated = PE - KE = 44,319.2 J - 14,058.58 J = 30,260.62 J

Therefore, the energy dissipated by friction on the ball is approximately 30,260.62 J.