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Mass of block given μK and Force?

by Ace.
Tags: μk, block, force, mass
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Ace.
#1
Nov15-12, 05:02 PM
P: 52
1. The problem statement, all variables and given/known data

A student pushes a block with 240 N of force across a horizontal surface with a coefficient of kinetic friction of 0.4. The block accelerates at a rate of 0.88 m/s2. Find the mass of the block.

2. Relevant equations

μK = FK / FN
F=ma

3. The attempt at a solution

My first attempt was to figure out mass using F=ma

m = f/a
= 240 N / 0.88 m/s2
= 272.7 kg
but this is not taking into account the kinetic friction...

So what I did was

FK = μK x FN
= 0.4 x (272.7 kg x 9.8 m/s2
= 1069.1 N
I know I'm doing this wrong and I need to know how to derive at net force to find acceleration
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Doc Al
#2
Nov15-12, 05:14 PM
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Quote Quote by Ace. View Post
My first attempt was to figure out mass using F=ma

m = f/a
= 240 N / 0.88 m/s2
= 272.7 kg
but this is not taking into account the kinetic friction...
To use ƩF = ma you must include all the forces.

So what I did was

FK = μK x FN
= 0.4 x (272.7 kg x 9.8 m/s2
= 1069.1 N
You are using the incorrect answer from your first attempt.

Instead, just call the mass "m" and set up your equation using the net force. Then you can solve for the mass. What's the friction force in terms of "m"?
Ace.
#3
Nov15-12, 05:45 PM
P: 52
You are using the incorrect answer from your first attempt.

Instead, just call the mass "m" and set up your equation using the net force. Then you can solve for the mass. What's the friction force in terms of "m"?
μK = FK / FN
FN = FK / μK
mg = FK / μK
m = (FK / μK) / g

Like this?

Doc Al
#4
Nov15-12, 05:58 PM
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Mass of block given μK and Force?

Quote Quote by Ace. View Post
μK = FK / FN
FN = FK / μK
mg = FK / μK
m = (FK / μK) / g

Like this?
No. (That won't work, since FK is proportional to the mass, so your equation ends up being m = m.)

Use ƩF = ma. First work on the forces. There are two forces acting. What are they?
Ace.
#5
Nov15-12, 06:40 PM
P: 52
force applied of 240 N and kinetic friction

I can't find magnitude of kinetic friction because I cannot find Force normal.
Doc Al
#6
Nov15-12, 06:55 PM
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Quote Quote by Ace. View Post
force applied of 240 N and kinetic friction
Good.

I can't find magnitude of kinetic friction because I cannot find Force normal.
If the mass is "m", what's the weight?
Ace.
#7
Nov15-12, 07:07 PM
P: 52
Quote Quote by Doc Al View Post
Good.


If the mass is "m", what's the weight?
weight = 9.8m
Doc Al
#8
Nov15-12, 07:35 PM
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Quote Quote by Ace. View Post
weight = 9.8m
Good. So now you have the normal force. Keep going.
Ace.
#9
Nov15-12, 07:49 PM
P: 52
μK = FK / FN
FK = μK FN
FK= 0.4 x 9.8m

I have 2 unknowns?
Doc Al
#10
Nov16-12, 04:01 AM
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Quote Quote by Ace. View Post
μK = FK / FN
FK = μK FN
FK= 0.4 x 9.8m
Good.

I have 2 unknowns?
The only unknown is the mass, which is what you are trying to find.

Keep going. What is ƩF?
Ace.
#11
Nov16-12, 05:04 AM
P: 52
Fapplied = 0.88m


just guessing here:
Fnet = FA [fwd] + Ff [bwd]
Fnet = FA [fwd] - Ff [fwd]
Fnet = 0.88m - (0.4 x 9.8m)
Doc Al
#12
Nov16-12, 05:08 AM
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Quote Quote by Ace. View Post
Fapplied = 0.88m
The applied force is given as 240 N.

just guessing here:
Fnet = FA [fwd] + Ff [bwd]
Fnet = FA [fwd] - Ff [fwd]
This is good.
Fnet = 0.88m - (0.4 x 9.8m)
Fix this and then use Fnet = ma.
Ace.
#13
Nov16-12, 05:17 AM
P: 52
oh wow my bad...

Fnet = 240 - (0.4 x 9.8m)
= 240 - 2.92m
Still don't get what I need to do next?
Doc Al
#14
Nov16-12, 05:39 AM
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Quote Quote by Ace. View Post
oh wow my bad...

Fnet = 240 - (0.4 x 9.8m)
= 240 - 2.92m
Still don't get what I need to do next?
Apply Newton's 2nd law: Fnet = ma
Ace.
#15
Nov16-12, 05:50 AM
P: 52
240 - 2.92m = 0.88m
240 = 0.88m + 2.92m
240 = 3.8m
m = 63 kg
Doc Al
#16
Nov16-12, 05:58 AM
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Quote Quote by Ace. View Post
240 - 2.92m = 0.88m
240 = 0.88m + 2.92m
240 = 3.8m
m = 63 kg
That's the right idea. But I think you made a mistake earlier (probably a typo):

Quote Quote by Ace. View Post
Fnet = 240 - (0.4 x 9.8m)
= 240 - 2.92m
Fix that, resolve your equation, and you'll have the correct mass.
Ace.
#17
Nov16-12, 06:05 AM
P: 52
Fnet = 240 - (0.4 x 9.8m)
= 240 - 3.92m

240 - 3.92m = 0.88m
240 = 0.88m + 3.92m
240 = 4.8m
m = 50 kg

wohooooooo thanks a lot!
Doc Al
#18
Nov16-12, 06:18 AM
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Yay!


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