What is the average force exerted on the putty by the surface?

[*intheclouds*]

Homework Statement

A 0.43 kg piece of putty is dropped from a height of 2.8 m above a flat surface. When it hits the surface, the putty comes to rest in 0.22 s. What is the average force exerted on the putty by the surface?

Homework Equations

F\Deltat=\DeltaP
F\Deltat=mv_f-mv_i
\DeltaP=mv_f-mv_i

(I don't think it is a collision, but if it is...any of those equations also)

The Attempt at a Solution

2.8m=(.22s)vo+.5(-9.8)(.22s)²
vo=13.8m/s

(.22s)F=(.43kg)(0m/s)-(.43kg)(13.8m/s)
F=-27N

But I know this isn't right...I just don't know what I did wrong...

 

[asleight]

Homework Statement

A 0.43 kg piece of putty is dropped from a height of 2.8 m above a flat surface. When it hits the surface, the putty comes to rest in 0.22 s. What is the average force exerted on the putty by the surface? ...

Energy conservation to the surface, then kinematics.

U_0+K_0=U_f+K_f\rightarrow U_0=K_f, and

F=ma=m\Delta v/\Delta t...

 

[Saladsamurai]

the .22s is how long the putty takes to come to a stop AFTER it hits the ground, not the time it takes from when it was released.

How can you find the velocity of the putty JUST BEFORE it hits the ground?

HINT: Use one of the kinematic equations that does not involve time.

Casey

 

[*intheclouds*]

Am I going to use the equation:
v²=v₀²+2ax??
Then it would be 0=v₀²+2(-9.8)(2.8).
Vo= 7.4 m/s
Is that what you are looking for?

 

[asleight]

Am I going to use the equation:
v²=v₀²+2ax??
Then it would be 0=v₀²+2(-9.8)(2.8).
Vo= 7.4 m/s
Is that what you are looking for?

You're right for a start; here's my logic:

U_0=K_f

Because it starts at rest, 2.8m above the ground, we know:

U_0 = mgh = 11.8 J = mv^2/2 \rightarrow v = 7.41 m/s

Then,

v_f=v_i+at\rightarrow a=. Because force and acceleration are related, you can continue from here.

 

[Saladsamurai]

Am I going to use the equation:
v²=v₀²+2ax??
Then it would be 0=v₀²+2(-9.8)(2.8).
Vo= 7.4 m/s
Is that what you are looking for?

I am not sure why you are solving for V_0

The putty is dropped and thus V_0=0 you need to solve for V_f

which is the Velocity of the putty immediately before it hits the ground

You get the same number, but only by sheer luck.

 

[*intheclouds*]

Ok thank you...I think I am just going to wait and ask my teacher. =]

 

[LowlyPion]

Homework Statement

A 0.43 kg piece of putty is dropped from a height of 2.8 m above a flat surface. When it hits the surface, the putty comes to rest in 0.22 s. What is the average force exerted on the putty by the surface?

Homework Equations

F\Deltat=\DeltaP
F\Deltat=mv_f-mv_i
\DeltaP=mv_f-mv_i

(I don't think it is a collision, but if it is...any of those equations also)

The Attempt at a Solution

2.8m=(.22s)vo+.5(-9.8)(.22s)²
vo=13.8m/s

(.22s)F=(.43kg)(0m/s)-(.43kg)(13.8m/s)
F=-27N

But I know this isn't right...I just don't know what I did wrong...

You've got part of it right.

The Average Force will be given by the change in momentum divided by the time interval. Since the final v is 0 then all you need to do is figure the momentum

V² = 2*a*x = 2*(9.8)*(2.8 m)

 

[*intheclouds*]

wow that actually helped a lot! thank you!

 

[Saladsamurai]

The key to this problem was to break it up into two sub-problems.

By identifying correctly that you needed to use Impulse and Momentum, you knew that you needed to find the Velocity of the putty just before it hit the ground.

That Velocity could be found by using kinematics (the first sub-problem).

On a side note, whenever you see forces (or moments/torques) and time involved, chances are pretty good that you can use Impulse and Momentum equations.

Casey

 

[*intheclouds*]

I got it...and the computer said my answer is right, so I am good to go! Thanks to everyone who helped! =]