Rate of Deceleration & Magnitude of Force

In summary, the ball decelerates until it comes to a stop at time 0.012. The magnitude of the unbalanced force exerted by the floor is 5040 Newtons.
  • #1
drsponge
7
0
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?
 
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  • #2
welcome to pf!

hi drsponge! welcome to pf! :smile:

(try using the X2 button just above the Reply box :wink:)
drsponge said:
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 :redface:
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? :wink:
 
  • #3
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.
 
  • #4
hi drsponge! :smile:
drsponge said:
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 Ftotal = ma …

the normal force N from the floor,

and the weight W = mg from the Earth :wink:
 
  • #5
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.
 
  • #6
yes :smile:

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 :wink:
 
  • #7
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.
 

Related to Rate of Deceleration & Magnitude of Force

1. What is the definition of rate of deceleration?

The rate of deceleration is the rate at which an object's velocity decreases over time. It is the change in velocity divided by the time it takes for that change to occur.

2. How is rate of deceleration calculated?

Rate of deceleration is calculated by dividing the change in velocity by the time it takes for that change to occur. The formula is: a = (v2-v1)/t, where "a" is the rate of deceleration, "v2" is the final velocity, "v1" is the initial velocity, and "t" is the time interval.

3. What factors affect the rate of deceleration?

The rate of deceleration is affected by several factors, including the initial velocity, the mass of the object, and the presence of external forces such as friction or air resistance.

4. How does the magnitude of force affect the rate of deceleration?

The magnitude of force directly affects the rate of deceleration. The greater the force applied to an object, the faster it will decelerate. This is because a greater force results in a larger change in velocity over a given time interval.

5. Is there a relationship between rate of deceleration and magnitude of force?

Yes, there is a direct relationship between rate of deceleration and magnitude of force. As the magnitude of force increases, the rate of deceleration also increases. This relationship can be described by Newton's Second Law of Motion: the acceleration of an object is directly proportional to the net force acting on it, and inversely proportional to its mass.

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