Question

Alice holds a black belt in Taekwondo and her fist has a mass of 0.5 kg. Her fist obtains a velocity of 5 m/s in 0.1 seconds from rest. Evaluate the average net force applied to the fist.

Answer 1

The average force on the fist is 25 N

Explanation:

First of all, we need to calculate the acceleration of the fist, which is equal to the rate of change of its velocity:

$a=\frac{v-u}{t}$

where:

u = 0 is the initial velocity (it starts from rest)

v = 5 m/s is the final velocity

t = 0.1 s is the time elapsed

Substituting,

$a=\frac{5-0}{0.1}=50 m/s^2$

Now we can calculate the average force applied to the fist by using Newton's second law of motion, which states that the net force applied on an object is equal to the product between its mass and its acceleration:

$F=ma$

where here we have:

m = 0.5 kg (mass of the fist)

$a=50 m/s^2$ (acceleration)

And solving for F, we find the force:

$F=(0.5)(50)=25 N$

Learn more about force and acceleration:

brainly.com/question/3820012

#LearnwithBrainly