average force physics help!?

Question

alex throws a .15kg ball down onto the floor. the ball's speed just before impact is 6.5m/s, and just after is 3.5m/s. if the ball is in contact with the floor for .025s, what is the magnitude of the average force applied by the floor on the ball?

how do i figure this out??

Answer 1

The average force applied by the floor on the ball is 60N upwards.

Newton's 2nd law of Motion is F = ma, but if you integrate it over a time interval during which the force varies, you can represent the effect of the average force ON THE BALL, Fbar(ball), by :

Fbar(ball) [delta t] = m [delta V]. In this case, measuring the average force AND the velocities UPWARDS,

Fbar(ball) [0.025s] = m [Vf - Vi] = 0.15 [3.5 - (- 6.5)] kg m/s ###

= 0.15 (10.0) kg m/s, so that

Fbar(ball) = (0.15/0.025) (10.0) kg m/s^2 = 60 N UPWARDS.

QED

[ Conversely, since action and reaction are equal and opposite, the average DOWNWARD force on the floor, should you be interested in that also, is given by

Fbar(floor) = - Fbar(ball) = 60N DOWNWARDS. ]

Live long and prosper.

### Note that ' SOMEBODY,' in the answer immediately preceding mine, unfortunately FORGOT that the velocities are REVERSED by the impact. That is why the magnitude of the force that he found was so small --- the correct CHANGE in speed, 10m/s was replaced by only 3m/s in his solution. Therefore his answer (18N) was only 3/10 of the correct answer (60N).

Answer 2

because of the fact the vehicle starts at 30 m/s then ends a 0 m/sec the rate is -30/m/sec (which potential the vehicle is decelerating not accelerating). Dividing the rate by way of the time required for the vehicle to end = -30 m/sec / 4.9 sec = -6.12 m/sec^2. using the equation F=ma the rigidity exerted to end the vehicle = ma = (.8 x 10^3 kg) x (-6.12 m/sec^2) = -4896 kg m/sec ^2 or -4896 N.

Answer 3

I would listen to Dr. Spock. He is both intelligent and logical. He also takes into account the change in direction in the vector of velocity, something the first responder does not do. Spock is right, as always, in this problem.