WILL GIVE BRAINLIEST A 225 g hockey puck is sliding on ice in an arena towards the end boards that are 15.7 m away. The puck is travelling 12.0 m/s when it slides into some rough ice (coefficient of kinetic friction= 0.550).
Determine:
a) the acceleration of the puck on the rough ice.
b) the distance from the end boards the puck is when it comes to a stop.

Please show work.

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j

Ray 2 should cross through the focal point on the right side of the lens