Mass attached to a spring

  • #1
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Homework Statement


I have a physics midterm on Wednesday and I'm doing some practice problems to prepare. On one of them, I'm getting an answer that is extremely close to the answer my TA gave me for the problem and I can't tell if it's rounding or if I'm missing something.

A mass is attached to a spring and allowed to move horizontally along a surface with friction (assume that the equilibrium position is far enough away from the wall that the mass never hits it and the spring can both push and pull). m=10kg and uK=uS=0.1

If the mass is let go, at x=70 cm (where x=0 is equilibrium) where does it come to rest for the first time?


Homework Equations


A) 1/2kx^2=1/2mv^2 -Ff (conservation of energy)
B) Ff=Nuk
C) F=-kx
D) v^2-vi^2=2a(x-xi)

The Attempt at a Solution


-I used equation B to find the frictional force: (0.1)*(9.8)*(10) = 9.8 N

-I used equation C to find the force exerted by the spring: F=(-100)(.7)=-70N. I then used F=ma to get the acceleration: (-70)=(10)a, and a=7m/s^2

I solved equation A for v and got squareroot((kx^2+2f)/m) = v
--after plugging and chugging, i got a velocity v=2.62 m/s

I then plugged into equation D

(2.62)^2-0=2(7)(x-0)
x=.49 m = 49 cm

The answer my TA gave me was 50.1 cm. Am I doing something incorrectly?

Thanks so much!
 

Answers and Replies

  • #2
haruspex
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Your equation A mixes energy and force.
The spring constant is 100N/m, right? I get -8.4cm.
 
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  • #3
rude man
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Homework Statement


I have a physics midterm on Wednesday and I'm doing some practice problems to prepare. On one of them, I'm getting an answer that is extremely close to the answer my TA gave me for the problem and I can't tell if it's rounding or if I'm missing something.

A mass is attached to a spring and allowed to move horizontally along a surface with friction (assume that the equilibrium position is far enough away from the wall that the mass never hits it and the spring can both push and pull). m=10kg and uK=uS=0.1

If the mass is let go, at x=70 cm (where x=0 is equilibrium) where does it come to rest for the first time?


Homework Equations


A) 1/2kx^2=1/2mv^2 -Ff (conservation of energy)
B) Ff=Nuk
C) F=-kx
D) v^2-vi^2=2a(x-xi)

The Attempt at a Solution


-I used equation B to find the frictional force: (0.1)*(9.8)*(10) = 9.8 N

-I used equation C to find the force exerted by the spring: F=(-100)(.7)=-70N. I then used F=ma to get the acceleration: (-70)=(10)a, and a=7m/s^2

I solved equation A for v and got squareroot((kx^2+2f)/m) = v
--after plugging and chugging, i got a velocity v=2.62 m/s

I then plugged into equation D

(2.62)^2-0=2(7)(x-0)
x=.49 m = 49 cm

The answer my TA gave me was 50.1 cm. Am I doing something incorrectly?

Thanks so much!
I got the same as the TA. except mine is negative: x = - 0.50m.
 
  • #4
haruspex
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I got the same as the TA. except mine is negative: x = - 0.50m.
That made me realise I hadn't read quite to the end of the question. My number was for where it comes to rest finally.
 
  • #5
ehild
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Homework Statement


I have a physics midterm on Wednesday and I'm doing some practice problems to prepare. On one of them, I'm getting an answer that is extremely close to the answer my TA gave me for the problem and I can't tell if it's rounding or if I'm missing something.

A mass is attached to a spring and allowed to move horizontally along a surface with friction (assume that the equilibrium position is far enough away from the wall that the mass never hits it and the spring can both push and pull). m=10kg and uK=uS=0.1

If the mass is let go, at x=70 cm (where x=0 is equilibrium) where does it come to rest for the first time?


Homework Equations


A) 1/2kx^2=1/2mv^2 -Ff (conservation of energy)
B) Ff=Nuk
C) F=-kx
D) v^2-vi^2=2a(x-xi)

The Attempt at a Solution


-I used equation B to find the frictional force: (0.1)*(9.8)*(10) = 9.8 N

-I used equation C to find the force exerted by the spring: F=(-100)(.7)=-70N. I then used F=ma to get the acceleration: (-70)=(10)a, and a=7m/s^2
You got the initial acceleration caused by the spring, and neglected the effect of the friction.
The acceleration is not constant during the motion, you can not use the kinematic equation D, valid for uniform acceleration.
 

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