Springs: Speed when released & distance traveled.

  • Thread starter Bradyns
  • Start date
  • #1
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Question:
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A 1.6 kg block is pushed into a spring with spring constant of 1.0 x 10^3 N/m. The spring is compressed a distance of 2.0 cm [0.02m], and the block is released from rest in the horizontal direction.

(a) Calculate the speed of the block as it passes through the equilibrium position at x = 0, if the surface is frictionless.

(b) How far does the block travel before coming to rest? Assume a constant friction force of 4.0 N and the same initial conditions as before.
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Formula I used for (a)

v = √[k/m (A^2 - x^2)]

v = √[1000/1.6 (0.02^2 - 0^2)]
v = 0.5m/s

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Not sure about (a), But I have no clue as to (b)..

Should I used the work formula?
Work (J) = Force x distance
 

Answers and Replies

  • #2
1,198
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Part a is correct.

Hint: Use Newton's second law followed by kinematic equation relating velocity, acceleration, and distance. Or use work energy relationship.
 
  • #3
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This should work for (b):

Conserving energy:
(1/2)×k x2 = f * d

where, k is spring constant, x is the compression of the spring, f is the frictional force, d is the distance traveled by the block before coming to a halt.
 
Last edited:
  • #4
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This should work for (b):

Conserving energy:
(1/2)×k x2 = f * d

where, k is spring constant, x is the compression of the spring, f is the frictional force, d is the distance traveled by the block before coming to a halt.

This gave me the same result [5cm, 0.05m] as one of the ways I tried it originally:

[v^2 = u^2 + 2ad]
Where:
a = F/m = 4N / 1.6kg = 2.5ms^-2
v = 0.5ms^-1
u = 0ms^-1

(0.5)^2 = (0) + 2 (2.5) d
0.25 = 5d
d = 0.25/5
d = 0.05m [5cm]

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Your method:
(1/2)×k x^2 = Ff * d
Where:
Ff = 4N
k = 1000
x = 0.02

(1/2)× (1000) (0.02^2) = 4 * d
500 x 0.0004 = 4 * d
0.2 = 4 * d
d = 0.2 / 4
d = 0.05m [5cm]
 
  • #5
58
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This gave me the same result [5cm, 0.05m] as one of the ways I tried it originally:

[v^2 = u^2 + 2ad]
Where:
a = F/m = 4N / 1.6kg = 2.5ms^-2
v = 0.5ms^-1
u = 0ms^-1

(0.5)^2 = (0) + 2 (2.5) d
0.25 = 5d
d = 0.25/5
d = 0.05m [5cm]

firstly, you put incorrect values into the formula v2=u2+2ad
v is the final velocity, u is the initial velocity. the final velocity has to be 0, as the block stops due to frictional force. secondly, you have taken the velocity that you obtained from (a), which is the velocity of the block if frictional force was not acting. also, the acceleration should be negative, because the block is decelerating due to the frictional force. also,

d = 0.25/5
d = 0.05m [5cm]

if you want to get the answer using your method, first find the velocity of the block where it leaves the spring, you can do this using energy conservation. now you can apply v2=u2+2ad
 

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