A60 kilogram student jumps down from a laboratory counter. at the instant he lands on the floor hus speed is 3 meters per second. if the student stops in .2 second what is thw average force of floor on the student

Newton's second law of motion states that an applied force is equal to the rate of change of momentum. This law is used to find out the magnitude of the force applied to the student.

The net force applied to the student is 900 N.

The net force is the vector sum of all the forces that act upon an object.

Given that a 60-kilogram student jumps down from a laboratory counter. His final velocity after 0.2 seconds is 3 m/s. The initial velocity of the student will be zero.

The net force acted on the student is calculated by Newton's second law of motion.

According to Newton's second law of motion, the rate of change of momentum of a body is equal to the resultant force acting on the body and is in the same direction.

Where F is the net force, P is the momentum, m is the mass, t is time, v is the final velocity and u is the initial velocity of the object.

Hence we can conclude that the net force applied to the student is 900 N.

To know more about Newton's second law of motion, follow the link given below.

link.

As per Newton's law rate of change in momentum is net force

so we can write it as

now we know that

from above equation

so he will experience 900 N force in above case

the inner core.

good luck: )

klicker