Acceleration coefficient

In summary, the problem involves finding the acceleration experienced by two masses (7kg and 12kg) on a plane with a coefficient of friction of .25. Using the equations for net force, it is determined that the net force on the two masses is 115.85 N. This results in an acceleration of 6.1 m/s2 for the 7kg mass and an acceleration of -6.1 m/s2 for the 12kg mass. The force of friction is calculated to be 1.75 N for the 7kg mass and the parallel force is determined to be 41.3 N.
  • #1
willingtolearn
62
0
Find the acceleration experienced by each of the two masses shown in the picture below if the coefficient of friction between the 7 kg mass and the plane is .25 ?
------------
In the picture, is the force of fiction go that way ?
http://img515.imageshack.us/img515/893/img3394dj7.jpg [Broken]
m1 = 7kg u = .25
m2 = 12kg
F (per) = F (normal)
F (per) = 68.6 cos 37 = 54.8 N
F (parallel) = 68.66 sin 37 = 41.3 N
F net = (12*9.8) - 1.75 = 115.85 N
115.85 / 19 = 6.1 m/s2
So 7 kg is accelerate at 6.1 m/s2 and 12 kg is the same but in the negative direction.
Is this right ?
 
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  • #2
can you explain this part:

"F net = (12*9.8) - 1.75 = 115.85 N"

how are you getting 1.75... and which mass are you taking Fnet for...
 
  • #3
learningphysics said:
can you explain this part:

"F net = (12*9.8) - 1.75 = 115.85 N"

how are you getting 1.75... and which mass are you taking Fnet for...

in this problem, there are only 2 force make the block able to slid. (and it is force of gravity on 12 kg and force friction on 7 kg.
So F net = F (gravity) - F (fiction)
 
  • #4
but how do you get 1.75N?

Also what about the F (parallel) = 68.66 sin 37 = 41.3 N?
 
  • #5
F (friction) = .25 time 7 = 1.75
 
  • #6
willingtolearn said:
F (friction) = .25 time 7 = 1.75

But friction = 0.25*Fnormal = 0.25*7*9.8*cos37.

Although it might be faster to do the whole problem at once... I think it's best to divide into 2 sections...

do the freebody diagram of the first mass, get the equations there... then the second mass, get the equations there... use T for tension...
 
  • #7
ok, I got this
a = 3.3 m/s2
 
  • #8
willingtolearn said:
ok, I got this
a = 3.3 m/s2

looks good to me. that's what I get also.
 
  • #9
Yes ... thanks learningphysics
 

1. What is an acceleration coefficient?

An acceleration coefficient is a measure of how much an object's velocity changes over a certain period of time. It is typically represented by the letter "a" and is expressed in units of distance per time squared.

2. How is acceleration coefficient calculated?

Acceleration coefficient is calculated by dividing the change in velocity by the change in time. This can be represented by the equation a = (vf - vi) / t, where vf is the final velocity, vi is the initial velocity, and t is the change in time.

3. What is the difference between acceleration and acceleration coefficient?

Acceleration refers to the change in velocity over time, while acceleration coefficient specifically refers to the numerical value of that change. In other words, acceleration is a concept, while acceleration coefficient is a measurement.

4. How does acceleration coefficient relate to Newton's Second Law of Motion?

According to 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. This means that a larger acceleration coefficient, or change in velocity, can be achieved with a greater force or a smaller mass.

5. What are some real-life examples of acceleration coefficient?

Acceleration coefficient can be observed in many everyday activities, such as riding a rollercoaster, driving a car, or throwing a ball. It is also a key concept in physics and engineering, used to analyze and design various systems, including rockets, airplanes, and sports equipment.

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