# Mass of block given μK and Force?

by Ace.
Tags: μk, block, force, mass
 P: 35 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 kgbut 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 NI know I'm doing this wrong and I need to know how to derive at net force to find acceleration
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P: 40,236
 Quote by Ace. My first attempt was to figure out mass using F=ma m = f/a= 240 N / 0.88 m/s2= 272.7 kgbut 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

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"?
P: 35
 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?

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P: 40,236

## Mass of block given μK and Force?

 Quote by Ace. 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?
 P: 35 force applied of 240 N and kinetic friction I can't find magnitude of kinetic friction because I cannot find Force normal.
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P: 40,236
 Quote by Ace. 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?
P: 35
 Quote by Doc Al Good. If the mass is "m", what's the weight?
weight = 9.8m
Mentor
P: 40,236
 Quote by Ace. weight = 9.8m
Good. So now you have the normal force. Keep going.
 P: 35 μK = FK / FN FK = μK × FN FK= 0.4 x 9.8m I have 2 unknowns?
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P: 40,236
 Quote by Ace. 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?
 P: 35 Fapplied = 0.88m just guessing here: Fnet = FA [fwd] + Ff [bwd] Fnet = FA [fwd] - Ff [fwd] Fnet = 0.88m - (0.4 x 9.8m)
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P: 40,236
 Quote by Ace. 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.
 P: 35 oh wow my bad... Fnet = 240 - (0.4 x 9.8m)= 240 - 2.92mStill don't get what I need to do next?
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P: 40,236
 Quote by Ace. oh wow my bad... Fnet = 240 - (0.4 x 9.8m)= 240 - 2.92mStill don't get what I need to do next?
Apply Newton's 2nd law: Fnet = ma
 P: 35 240 - 2.92m = 0.88m 240 = 0.88m + 2.92m 240 = 3.8m m = 63 kg
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P: 40,236
 Quote by Ace. 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 by Ace. Fnet = 240 - (0.4 x 9.8m)= 240 - 2.92m
Fix that, resolve your equation, and you'll have the correct mass.
 P: 35 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!
 Mentor P: 40,236 Yay!

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