A community for students.
Here's the question you clicked on:
 0 viewing
haridas_mandal
 2 years ago
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)
haridas_mandal
 2 years ago
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)

This Question is Closed

haridas_mandal
 2 years ago
Best ResponseYou've already chosen the best response.0Speed is 4.9 m/s . My mistake.

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0try 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.

haridas_mandal
 2 years ago
Best ResponseYou've already chosen the best response.0a =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?

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0what does "breaking strength 40 kgwt" mean??

haridas_mandal
 2 years ago
Best ResponseYou've already chosen the best response.0I 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
 2 years ago
Best ResponseYou've already chosen the best response.0I am not sure ... since no details have given.

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0is the length of wire given?

haridas_mandal
 2 years ago
Best ResponseYou've already chosen the best response.0No .I have posted the question verbatim from her question paper.

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.0try putting F = 40*9.8

haridas_mandal
 2 years ago
Best ResponseYou've already chosen the best response.0Then 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
 2 years ago
Best ResponseYou've already chosen the best response.0the problem is with "40 kgwt." I can't decipher what it means ... this is 11th grade Q right?

haridas_mandal
 2 years ago
Best ResponseYou've already chosen the best response.0Yeah 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
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1350837934080:dw

haridas_mandal
 2 years ago
Best ResponseYou've already chosen the best response.0If 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
 2 years ago
Best ResponseYou've already chosen the best response.0try opposite, using those distances, find acceleration (4m, 0.25m, 1 m or 1.25 m) calculate force and see how this relates to 40kgwt

haridas_mandal
 2 years ago
Best ResponseYou've already chosen the best response.0Ok i shall do that. but got to leave now . Shall catch up .Thanks.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.