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Mass-string system problem

  • #1

Homework Statement


A body of mass 5kg is suspended at the lower end of a mass-less string, which passing through a mass-less pulley and is being pulled down at the other end by a force of 500N.
The force of the pull reduces at a rate of 30N per meter, through which the body is raised.

What is the velocity of the body when it reaches a height 10m, starting from rest?
At what height will the direction of velocity change?


Homework Equations


a=dv/dt=v.dv/dx
T-mg=ma
T=f, where f = F-30y
so basically, F-mg=ma


The Attempt at a Solution


starting from F-mg=ma, i went to deriving:
(F-30y)-mg=m(dv/dt)=m.v.dv/dy
so, (F-mg)*y -15(y^2) = (m/2)v^2
=> (500-5*10)*10 - 15*(100) = (5/2)v^2
=> 4500 - 1500 = (5/2)v^2
=> 1200 = v^2

so v = 10√12 (that's my answer)

for the velocity to change direction, it must first become zero again - so putting zero as the initial and final values limits of integration, i got:
(F-mg)*y -15(y^2) = 0
=> [F -mg -15y]*y = 0

so y = 0 or (F-mg)/15 = 30

as y=0 is starting point, it cant be accepted as the required answer

hence, at y=30, the block will change it's velocity


IS WHAT I DID CORRECT?
 

Answers and Replies

  • #2
6,054
390
Everything looks good.

Note you could have obtained your results more "physically" by using work/energy explicitly.
 
  • #3
Can you please show me how to do that - just a short, brief set of instructions will do
 
  • #4
6,054
390
The change in the kinetic energy of the body is given by the work of the resultant force. This leads directly to the equation you obtained through integration.
 
  • #5
haruspex
Science Advisor
Homework Helper
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The change in the kinetic energy of the body is given by the work of the resultant force. This leads directly to the equation you obtained through integration.
But since the force is changing it will still require integration, no?
 
  • #6
1,540
134
zibs.shirsh...This is how you would approach using work energy method

Let T be the variable tension in the string .Displacement of mass is considered positive upwards.

ΔT/Δx=-30 N/m

Integrate with proper limits and you would get T as the function of x .

Now apply work energy theorem ,∫Tdx = ΔKE + ΔPE .
 
  • #7
784
11
so the fact that the question mentions a spring is just a red herring...?
 
  • #8
6,054
390
There is no spring in the question. There is a string.
 

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