aA player uses a hockeey stick to incease the speed of a 0.200 kg hockey puck by 6 m /s in 2 seconds.

(A)How much dd the hockey puck accelerate?
(B)How much force was exerted on the pack?
(C)How much force did the puck exert on the hockey stick.

To find the answers to these questions, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a): F = m * a.

(A) To find the acceleration of the hockey puck, we can use the formula: acceleration (a) = change in velocity (Δv) / time (t).

Given: change in velocity (Δv) = 6 m/s, time (t) = 2 s.
So, a = Δv / t = 6 m/s / 2 s = 3 m/s².
Therefore, the hockey puck accelerated at a rate of 3 m/s².

(B) To find the force exerted on the puck, we can use the formula: F = m * a.

Given: mass (m) = 0.200 kg, acceleration (a) = 3 m/s².
So, F = 0.200 kg * 3 m/s² = 0.600 N.
Therefore, the force exerted on the hockey puck is 0.600 Newtons.

(C) According to Newton's third law of motion, for every action, there is an equal and opposite reaction. So, the force exerted by the puck on the hockey stick is equal in magnitude but opposite in direction to the force exerted by the stick on the puck. Therefore, the force exerted by the puck on the hockey stick is also 0.600 Newtons.