A person with a black belt in karate has a fist that has a mass of 0.70 kg. Starting from rest, this fist attains a velocity of 6.9 m/s in 0.15 s. What is the magnitude of the average net force applied to the fist to achieve this level of performance?
F = dP/dt = 0.70(6.9-0)/0.15
= 32.2 N
37n
To find the magnitude of the average net force applied to the fist, we can use Newton's second law of motion, which states that the force applied to an object is equal to the mass of the object multiplied by its acceleration:
Force = Mass * Acceleration
First, let's find the acceleration of the fist. We can use the formula for acceleration:
Acceleration = (Final Velocity - Initial Velocity) / Time
Given:
Mass (m) = 0.70 kg
Initial Velocity (u) = 0 m/s (starting from rest)
Final Velocity (v) = 6.9 m/s
Time (t) = 0.15 s
Plugging these values into the formula, we get:
Acceleration = (6.9 m/s - 0 m/s) / 0.15 s
Simplifying, we have:
Acceleration = 6.9 m/s / 0.15 s
Now, let's calculate the acceleration:
Acceleration = 46 m/s²
Now that we have the acceleration, we can find the magnitude of the average net force. Plugging the values into Newton's second law:
Force = Mass * Acceleration
Force = 0.70 kg * 46 m/s²
Calculating the force, we get:
Force = 32.2 N
Therefore, the magnitude of the average net force applied to the fist is 32.2 Newtons.