Friction produces heat hope this helps
The answer is 4.0 kg since the flywheel comes to rest the
kinetic energy of the wheel in motion is spent doing the work. Using the
formula KE = (1/2) I w².
Given the following:
I = the moment of inertia about the
axis passing through the center of the wheel; w = angular velocity ; for the
solid disk as I = mr² / 2 so KE = (1/4) mr²w². Now initially, the wheel is spinning
at 500 rpm so w = 500 * (2*pi / 60) rad / sec = 52.36 rad / sec.
The radius = 1.2 m and KE = 3900 J
3900 J = (1/4) m (1.2)² (52.36)²
m = 3900 J / (0.25) (1.2)² (52.36)²
m = 3.95151 ≈ 4.00 kg
consider the velocity of the ball towards the wall as negative and away from the wall as positive.
m = mass of the ball = 513 g = 0.513 kg
v₀ = initial velocity of the ball towards the wall before collision = - 14.7 m/s
v = final velocity of the ball away from the wall after collision = 11.3 m/s
t = time of contact with the wall = 0.038 sec
F = average force acting on the ball
using impulse-change in momentum equation , average force is given as
F = m (v - v₀)/t
inserting the values
F = (0.513) (11.3 - (- 14.7))/0.038
F = 351 N
I think it’s C) lmk if I’m wrong
