Spring launching 3 kg block

  • Thread starter jjd101
  • Start date
  • #1
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Homework Statement


A horizontal spring with a spring constant of 200.0 N/m is compressed 25.0 cm and used to launch a 3.00 kg block across a frictionless surface. After the block travels some distance, the block goes up a 32 degree incline that has a coefficient of friction of 0.25 between the block and the surface of the incline. How far along the incline does the block go before stopping?



Homework Equations


motion equations
f=-ks
f=ma


The Attempt at a Solution


i tried taking the force of the spring at .25m compression which is 50N times sin32 to account for incline i got 26.5N, minus the force of the friction which i got as 6.23 so a total force of 20.27 divided by the mass gives acceleration of 6.76. Then i was not able to use a motion equation to find a distance up the incline
 

Answers and Replies

  • #2
998
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... i tried taking the force of the spring at .25m compression which is 50N times sin32 to account for incline i got 26.5N,...

When the mass is on the incline there is no component of the force of the spring anymore. The effect of the spring appears only in the velocity of the mass up the plane.

Which forces do you think act on the mass while it is moving up the incline?
 
  • #3
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force of friction and gravity?
 
  • #4
998
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There is also the normal reaction of the incline on the mass.

Now which method are you going to use, energy or equations of motion?
 
  • #5
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okay
 
  • #6
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it would be mgsin32 i believe
 
  • #7
998
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the normal reaction is mgcos32.
 
  • #8
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so normal incline is mgcos32 = 24.93 N and ff = .25mg =7.35N so total resistance along the incline is 32.28N?
 
  • #9
998
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normal reaction = 0.25x24.93=6.23N
 
  • #10
998
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frictional force = coefficient of friction x normal reaction
 
  • #11
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thats what i got originally, how do i use that to find a distance up the ramp?
 
  • #12
998
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are you going to use energy or equations of motion?
 
  • #13
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im guessing energy because i didnt see a way of doing it with motion equations?
 
  • #14
998
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so initial energy = final energy
what is formula for the initial energy?
what is formula for final energy?
 
  • #15
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initial energy = Us=1/2kX^2 = 1/2(200)(.25)^2 = 6.25 KE=1/2mv^2 6.25=1/2(3)V^2
Vi= 2.04?
 
  • #16
998
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correct.
In what form is the final energy?
 
  • #17
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Kinetic Energy with a velocity 0? or maybe potential energy cause its not moving?
 
  • #18
998
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if v = 0 there is not ke
So final energy is mgh.

Hence can you write an equation showing what happened to the intial energy?
 
  • #19
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so 6.25=mgh so h= .213m ?
 
  • #20
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6.25J has to provide energy for the final PE (mgh) and ALSO for the work done against friction.
 
  • #21
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how do i find the work done by friction?
 
  • #22
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initial energy = PE + work against friction

6.25 = mgh + frictional force x distance along incline

6.25 = mg x distance along incline x sin32 + frictional force x distance along incline
 
  • #23
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i just did that calculation and i got .0314m along the incline?
 
  • #24
998
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i got 0.286m
 

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