Finding the Accelation Of A Body In Presence Of Frictions

In summary: In this case, the first statement is that the force is just sufficient to get the body moving in the first place. What does this tell you about the force with which you are pushing the body?In summary, a 60 kg block is pushed horizontally with just enough force to start its motion Accross a floor and the same force acts on it afterwards. The coefficient of static and sliding frictions are 0.5 and 0.4 respectively. Find acceleration of the body.
  • #1
Navin
112
34
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1. Homework Statement

A 60 kg block is pushed horizontally with just enough force to start its motion Accross a floor and the same force acts on it afterwards.The coefficient of static and sliding frictions are 0.5 and 0.4 respectively.
Find acceleration of the body.
Options

6m/s^2

4.9m/s^2

3.92m/s^2

1m/s^2

Homework Equations


F= ma

Friction = (u)m.a where " u " is coefficient of friction

The Attempt at a Solution


First I assumed that the total force need would be the summation of all the friction forces acting on the body

So I did
m.a = friction static + friction kinetic

60. a = 540

a = 9 m/s^2
But alas tis the wrong answer.

Then I tried another method which is in the attachment above I hope it is visible. ...but I'm pretty sure it's wrong.

[So don't delete the post oh ,moderator supreme overlord for I have tried !]

I would really like to know how to approach this problem

THIS IS ALL THE INFO GIVEN
 

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  • #2
The static coefficient of friction gives you the maximal possible friction force when the object is stationary. The kinetic coefficient of friction gives you the friction when the object is moving. They will not both be relevant at the same time.
 
  • #3
Orodruin said:
The static coefficient of friction gives you the maximal possible friction force when the object is stationary. The kinetic coefficient of friction gives you the friction when the object is moving. They will not both be relevant at the same time.
So when I make the equation for in motion I should just consider kinetic right ?

and secandly since the force is constant will there be change in accelaration to that of when it was set into motion ?
 
  • #4
Think about how you will use each piece of information. (Although there is one piece of information that is superfluous, but let's find out which. The first statement is that the force is just sufficient to get the body moving in the first place. What does this tell you about the force with which you are pushing the body?
 
  • #5
Orodruin said:
Think about how you will use each piece of information. (Although there is one piece of information that is superfluous, but let's find out which. The first statement is that the force is just sufficient to get the body moving in the first place. What does this tell you about the force with which you are pushing the body?
Well i think it tells us that the force is just sufficient to tip the balance.like its at the Citical value...therefore I think it's right to assume(please correct me if I'm wrong)
That the force will be equal to static friction
Hence
F=Friction (static)
F = 300
 
  • #6
Navin said:
That the force will be equal to static friction
Yes, this is correct. The force is just sufficient to overcome static friction, which in effect is saying that it is essentially equal to the maximal static friction.

Navin said:
Hence
F=Friction (static)
F = 300
Units, please! Units are important when you do physics. If you do not use them, you will have a difficult time. Even better in this case would be to leave the symbolic expression for the force and only input numbers at the very end. This is often (very) helpful in understanding and finding errors. In this case, what is the symbolic expression for the force ##F## expressed in the mass ##m##, the gravitational acceleration ##g##, and the coefficient of static friction ##\mu_s##?
(If you are unfamiliar with LaTeX, you can use the symbols available when you press the ##\Sigma## in the editor to get μ and the ##x_2## symbol to make a subindex s, all in all giving μs.)
 
  • #7
↔↔
Orodruin said:
Yes, this is correct. The force is just sufficient to overcome static friction, which in effect is saying that it is essentially equal to the maximal static friction.Units, please! Units are important when you do physics. If you do not use them, you will have a difficult time. Even better in this case would be to leave the symbolic expression for the force and only input numbers at the very end. This is often (very) helpful in understanding and finding errors. In this case, what is the symbolic expression for the force ##F## expressed in the mass ##m##, the gravitational acceleration ##g##, and the coefficient of static friction ##\mu_s##?
(If you are unfamiliar with LaTeX, you can use the symbols available when you press the ##\Sigma## in the editor to get μ and the ##x_2## symbol to make a subindex s, all in all giving μs.)
Great ! I think I got the answer (hopefully) the working is in the attachment

I'm sorry for not using units,I'll make sure to use them henceforth.im extremely unfamiliar with the keyboard layout !
 

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  • #8
Please stop attaching pictures instead of typing your work. It violates forum policy.

Also, if you used symbolic expressions all the way rather than inserting numbers in every step, you would realize which piece of information in the problem was superfluous.
 
  • #9
Orodruin said:
Please stop attaching pictures instead of typing your work. It violates forum policy.

Also, if you used symbolic expressions all the way rather than inserting numbers in every step, you would realize which piece of information in the problem was superfluous.
Gosh I should really read the terms and conditions ! Sorry for the attachments

mass is superfluous isn't it ?
 
  • #10
Navin said:
mass is superfluous isn't it ?
Yes. All of the forces are proportional to the mass, which means it cancels out with the ##m## in Newton's second law.
 
  • #11
Orodruin said:
Yes. All of the forces are proportional to the mass, which means it cancels out with the ##m## in Newton's second law.
Thank you so much ! This problem has been bugging me for a while .

Anyway have an awesome day !
 

1. What is acceleration in the presence of friction?

Acceleration in the presence of friction is the rate at which the velocity of a body changes due to the opposing force of friction. It is a measure of how quickly an object speeds up or slows down in response to the frictional force acting upon it.

2. How is acceleration affected by friction?

Friction can cause the acceleration of a body to decrease, since it acts in the opposite direction of the applied force. This means that the body will take longer to reach a certain velocity or change direction, compared to if there was no friction present.

3. What factors affect the acceleration of a body in the presence of friction?

The acceleration of a body in the presence of friction is affected by the magnitude of the applied force, the mass of the object, and the coefficient of friction between the surfaces in contact. Other factors such as surface roughness and temperature can also play a role.

4. How is the coefficient of friction determined?

The coefficient of friction is determined experimentally by measuring the force required to move an object across a surface at a constant speed. The ratio of this force to the normal force acting on the object is equal to the coefficient of friction.

5. How can we calculate the acceleration of a body in the presence of friction?

The acceleration of a body in the presence of friction can be calculated using Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. The net force is the difference between the applied force and the force of friction, which can be determined using the coefficient of friction and the normal force.

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