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?
PE = mgh = 28*9.8*159
KE = 1/2 mv^2 = 1/2 * 28 * 31.3^2
Because of friction, not all of the PE changed to KE. Subtract to see how much energy was lost.
To determine the amount of energy dissipated by friction on the ball, we need to analyze the change in kinetic energy of the ball as it rolls down the hill.
The initial kinetic energy (KE_initial) of the ball when it is at rest is zero because it is not moving. The final kinetic energy (KE_final) of the ball when it reaches the bottom of the hill is given by the formula:
KE_final = (1/2) * m * v^2
where m is the mass of the ball (28 kg) and v is the final velocity of the ball (31.3 m/s).
Using the given values, we can calculate the final kinetic energy:
KE_final = (1/2) * 28 kg * (31.3 m/s)^2
= 14 kg * (980.09 m^2/s^2)
= 13721.26 J
Now, let's calculate the difference in kinetic energy between the initial and final states:
∆KE = KE_final - KE_initial
= 13721.26 J - 0 J
= 13721.26 J
This difference in kinetic energy represents the energy dissipated by friction on the ball as it rolls down the hill. Therefore, the amount of energy dissipated by friction on the ball is 13721.26 Joules.