In attempting to pass the puck to a teammate, a hockey player gives it an initial speed of 1.43 m/s. However, this speed is inadequate to compensate for the kinetic friction between the puck and the ice. As a result, the puck travels only one-half the distance between the players before sliding to a halt. What minimum initial speed should the puck have been given so that it reached the teammate, assuming that the same force of kinetic friction acted on the puck everywhere between the two players?
I tried solving this by using 1/2Vo^2 but it comes out incorrect. What am I missing?
2 answers
I wouldn't know how to solve the problem either because you are missing some information.
I apologize about that. I should have thought about it for more than 3 seconds. The answer is in the directions: the puck travels only one-half the distance
Conservation of mechanical energy says that
Work=kinetic energy
F*d=1/2mv^2
m*a*d=1/2mv^2
masses are the same and cancel.
a*d=1/2v^2
Solving for a (acceleration):
a=[1/2v^2]/d
I need to increase d by 2, and I can only increase the velocity, per the directions
In both situations, acceleration will be the same. So, how much would I need to increase v by to increase d (displacement) by 2?
a=[1/2v^2]/d
[1/2(1.43m/s)^2]d=[1/2v^2]2d
(1.43m/s)^2/d=v^2/2d
(2.04m^2/s^2)/d=v^2/2d
2*(2.04m^2/s^2)=v^2
4.08m^2/s^2=v^2
Sqrt*[4.08m^2/s^2]=v
v=2.02m/s
Conservation of mechanical energy says that
Work=kinetic energy
F*d=1/2mv^2
m*a*d=1/2mv^2
masses are the same and cancel.
a*d=1/2v^2
Solving for a (acceleration):
a=[1/2v^2]/d
I need to increase d by 2, and I can only increase the velocity, per the directions
In both situations, acceleration will be the same. So, how much would I need to increase v by to increase d (displacement) by 2?
a=[1/2v^2]/d
[1/2(1.43m/s)^2]d=[1/2v^2]2d
(1.43m/s)^2/d=v^2/2d
(2.04m^2/s^2)/d=v^2/2d
2*(2.04m^2/s^2)=v^2
4.08m^2/s^2=v^2
Sqrt*[4.08m^2/s^2]=v
v=2.02m/s