Avery hard rubber ball (m = 0.5 kg) is falling vertically at 4 m/s just before it bounces on the floor. the ball rebounds back at essentially the same speed. if the collision with the floor lasts 0.05 s, what is the average force exerted by the floor on the ball?

The force exerted by the floor is 80 N.

Explanation:

Given that,

Mass of ball = 0.5 kg

Velocity= 4 m/s

Time t = 0.05 s

When the ball rebounds then the kinetic energy is

Where, m = mass of ball

v = velocity of ball

Put the value into the formula

The average force exerted by the floor on the ball = change in kinetic energy over collision time

Hence, The force exerted by the floor is 80 N.

answer: a. same unit

explanation:

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.