Question

In attempting to pass the puck to a teammate, a hockey player gives it an initial speed of 1.94 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?

can you also please do the same question but use 1.02 m/s so,

In attempting to pass the puck to a teammate, a hockey player gives it an initial speed of 1.02 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?

This would help so much if you could answer both 1.94 and 1.02 m/s.. THANKS!

Answer 1

When vi = 1.94 m/s

How much kinetic energy does the puck have initially?

KE = 1/2 m vi^2 = 1/2 * (1.94)^2 * m = 1.88*m J

Work is done by friction to slow the puck down. If the puck stops, all of its kinetic energy has been lost due to the work done by friction. Work can be related to a constant force over a distance.

W = F * d =1.88*m J

Since I must travel twice the distance with the same constant frictional force, friction must do twice as much work and my puck must have twice as much kinetic energy.

1.88*m*2 = 1/2*m*v^2

1.88*2*2 = v^2

(7.527)^(1/2) = v =2.74 m/s....Answer.

When Vi = 1.02m/s

How much kinetic energy does the puck have initially?

KE = 1/2 m v^2 = 1/2 * (1.02)^2 * m = 0.520*m J

Work is done by friction to slow the puck down. If the puck stops, all of its kinetic energy has been lost due to the work done by friction. Work can be related to a constant force over a distance.

W = F * d =0.520*m J

Since I must travel twice the distance with the same constant frictional force, friction must do twice as much work and my puck must have twice as much kinetic energy.

0.520*m*2 = 1/2*m*v^2

0.520*2*2 = v^2

(2.080)^(1/2) = v =1.44m/s....Answer.

Hope this helps you.