Rate of Deceleration & Magnitude of Force

[drsponge]

I'm not entirely sure of my answer, if it's wrong could some one point me in the right direction?

Homework Statement

A 140g ball is falling through the air at 36 meters per second (constant speed) prior to time 0.
The ball impacts the floor at time 0. The ball decelerates until time 0.012. At this time ball has come to rest.

Between time 0 and time 0.012 what is the rate of deceleration?

Then using this data calculate the magnitude of the force exerted by the floor.

Homework Equations

Deceleration = initial speed minus final speed divided by time.
36 m s-1 - 0 m s-1 / 0.012 s
= 3000 m s -1

Calculate the magnitude of the unbalanced force exerted by the floor.
F = ma
F = 140g x 36 m s-1 = 5040 Newtons.

OR IS IT

F = 140g x 3000 m s-1 = 420,000 Newtons.

The Attempt at a Solution

The solution would be 5040 Newtons was being exerted by the floor. But is this correct? I under stand that acceleration is a CHANGE in velocity with in time. If the speed is constant until there is no acceleration only deceleration after the impact?

 

[tiny-tim]

welcome to pf!

hi drsponge! welcome to pf!

(try using the X² button just above the Reply box )

Deceleration = initial speed minus final speed divided by time.
36 m s-1 - 0 m s-1 / 0.012 s
= 3000 m s -1

you mean 3000 m s^-2

Calculate the magnitude of the unbalanced force exerted by the floor.
F = ma
F = 140g x 36 m s-1 = 5040 Newtons.

OR IS IT

F = 140g x 3000 m s-1 = 420,000 Newtons.

i] Newton is an SI (kgs) unit, so you must convert all masses to kg

ii] it's 3000 m s^-2 …

so the question is, should you be using m s^-1 or m s^-2 ?​

iii] what about gravity?

 

[drsponge]

Oh yes, s^-2 would be correct.

The ball weighs 140g or 0.14kg. It is traveling at a constant speed of 36 m s^-1 until impact with the floor at point 0.

The change in acceleration is 36 m s^-1 minus 0 m s^-1 over 0.012 s which means the change in acceleration is 3000 m s^-2

Therefore the force exerted by the floor is
F = 0.14kg x 3000 m s^-2
= 420 N

In regards with gravity if the ball is falling with a constant velocity it must have reached it's terminal velocity so gravity and drag / air resistance are equal before impact.

 

[tiny-tim]

hi drsponge!

In regards with gravity if the ball is falling with a constant velocity it must have reached it's terminal velocity so gravity and drag / air resistance are equal before impact.

yes, but drag disappears when the ball hits the floor, while gravity doesn't disappear!

there are two forces on the ball, that make up F_total = ma …

the normal force N from the floor,

and the weight W = mg from the Earth ​

 

[drsponge]

Thanks Tim,

Ahh so the net force is F₂ because W = mass (0.14kg) x gravity (9.8 m s²) is acting on the ball and so is the force exerted by the floor in N.

W = 1.37kg
N = 420

Since 1 N = 1 kg m s^-2

Therefore the forces are unbalanced because the floor is exerting more force than gravity which halts the balls progress.

 

[tiny-tim]

yes

since we're using two significant figures, it makes no difference, but if we were using three, then the normal force would be 421 N

 

[drsponge]

Hi Tim,

Thanks very much for all your help. I think I've managed to safely file this away in my brain. I'm starting out on a science qualification towards a degree eventually. First time I've even attempted science academically for 10 years or so.. I'm rusty.