anonymous
  • anonymous
A 20kg body is connected to a rope of breaking strength 40 kgwt. If it is moved down with a speed of 9.5m/s at certain instant. The shortest distance in which it can be stopped is (4m, 0.25m, 1 m or 1.25 m)
Physics
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Speed is 4.9 m/s . My mistake.
experimentX
  • experimentX
try to find acceleration using F=ma ... probably F=40 and m=20 then use this formula v^2 = u^2 + 2 a s u=0, v=4.9, use 'a' from above ... find 's' which is distance.
anonymous
  • anonymous
a =40/20=2 m/s^2. hence S = 4.9 x4.9 /2x2 =6.0025 .but options are 4,0.25,1 &1.25. How to go about it?

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experimentX
  • experimentX
what does "breaking strength 40 kgwt" mean??
anonymous
  • anonymous
I am not very sure .I am trying help my daughter in her homework. As I understand it should the maximum force which that rope can withstand.Will there be force due to gravity play its part?
experimentX
  • experimentX
I am not sure ... since no details have given.
experimentX
  • experimentX
is the length of wire given?
anonymous
  • anonymous
No .I have posted the question verbatim from her question paper.
experimentX
  • experimentX
try putting F = 40*9.8
anonymous
  • anonymous
Then a = 40x9.8/20 =19.6 m/sec^2 & S = 4.9 x4.9/2x 19.6=0.6125. But it is not near to any of the four options given.
experimentX
  • experimentX
the problem is with "40 kgwt." I can't decipher what it means ... this is 11th grade Q right?
anonymous
  • anonymous
Yeah it is class 11 std in India. My daughter is going through a recap as she is to sit for a National level exam for admission to med college.
experimentX
  • experimentX
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anonymous
  • anonymous
If we take initial vel as 4.9 m/s and allow force of gravity then v^2 =4.9^2 +2x 9.8x S or Shortest dist =( v^2 -9.8^2)/9.8x2 but we do not know the value of V final. Can there be some other way?
experimentX
  • experimentX
try opposite, using those distances, find acceleration (4m, 0.25m, 1 m or 1.25 m) calculate force and see how this relates to 40kgwt
anonymous
  • anonymous
Ok i shall do that. but got to leave now . Shall catch up .Thanks.

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