a player uses a hockey stick to increase the speed of a 0.200 kg hockey puck by 6 m/s in 2 seconds.
how much force was exerted on the puck?
how much force did the puck exert on the hockey stick?
Force = rate of change of momentum (in other words m (delta v/ delta t) or m a)
F = m (change in velocity/change in time)
F = .2 (6/2) = .6 Newtons average during the 2 seconds.
Another way, which is really the same, is to say Force * time = impulse which is mass times change in velocity
F * 2 = .2 (6)
F = .6 Newtons again
Newton's third law says same force, .6 Newtons exerted back on the stick by the puck
To calculate the force exerted on the puck and the force exerted on the hockey stick, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) times acceleration (a).
1. Force exerted on the puck:
Given:
- Mass of the puck (m): 0.200 kg
- Change in velocity (Δv): 6 m/s
- Time (t): 2 seconds
First, we need to determine the acceleration of the puck using the formula a = Δv / t:
Acceleration (a) = Δv / t = 6 m/s / 2 s = 3 m/s^2
Now, we can calculate the force exerted on the puck using the formula F = m * a:
Force (F) = m * a = 0.200 kg * 3 m/s^2 = 0.6 N
Therefore, the force exerted on the puck is 0.6 Newtons.
2. Force exerted on the hockey stick:
According to Newton's third law of motion, every action has an equal and opposite reaction. Therefore, the force exerted on the hockey stick by the puck will be the same magnitude but in the opposite direction as the force exerted on the puck. So, the force exerted on the hockey stick will also be 0.6 Newtons, but in the opposite direction.
In summary:
- Force exerted on the puck: 0.6 N in the direction of motion.
- Force exerted on the hockey stick: 0.6 N in the opposite direction of motion.