A) a = (V2-V1)/time = (6.0 m/s)/2 s
= 3.0 m/s^2
B) F = M*a (on puck, forward)
Call this the 'action' force
C) reaction = -action
is the force on the stick
(Newton's third law)
Do the numbers.
A player uses a hockey stick to increase the speed of a 0.200kg hockey puck by 6 m.s in 2 seconds.
A) How much did the hockey puck accelerate?
B) How much force was exerted on the puck?
C) How much force did the puck exert on the stick?
1 answer