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?
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.
To answer these questions, we can use Newton's second law of motion, which states that force (F) is equal to the mass (m) of an object multiplied by its acceleration (a). The formula can be written as F = m * a.
A) To find the acceleration of the hockey puck, we can use the formula for acceleration: a = (final velocity - initial velocity) / time. Here, the initial velocity is 0 m/s (because the puck was at rest), the final velocity is 6 m/s, and the time is 2 seconds. Plugging these values into the formula, we have:
a = (6 m/s - 0 m/s) / 2 s
a = 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 Newton's second law. We know the mass of the puck is 0.200 kg and the acceleration is 3 m/s². Plugging these values into the formula, we have:
F = m * a
F = 0.200 kg * 3 m/s²
F = 0.600 N
Therefore, the force exerted on the puck is 0.600 Newtons.
C) Newton's third law of motion states that for every action, there is an equal and opposite reaction. Therefore, the force exerted by the puck on the stick is equal in magnitude but opposite in direction to the force exerted by the stick on the puck. So, the force exerted by the puck on the stick is also 0.600 Newtons, but in the opposite direction.