Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
haridas_mandal
Group Title
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)
 one year ago
 one year ago
haridas_mandal Group Title
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)
 one year ago
 one year ago

This Question is Closed

haridas_mandal Group TitleBest ResponseYou've already chosen the best response.0
Speed is 4.9 m/s . My mistake.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

haridas_mandal Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
what does "breaking strength 40 kgwt" mean??
 one year ago

haridas_mandal Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
I am not sure ... since no details have given.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
is the length of wire given?
 one year ago

haridas_mandal Group TitleBest ResponseYou've already chosen the best response.0
No .I have posted the question verbatim from her question paper.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
try putting F = 40*9.8
 one year ago

haridas_mandal Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
the problem is with "40 kgwt." I can't decipher what it means ... this is 11th grade Q right?
 one year ago

haridas_mandal Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
dw:1350837934080:dw
 one year ago

haridas_mandal Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

haridas_mandal Group TitleBest ResponseYou've already chosen the best response.0
Ok i shall do that. but got to leave now . Shall catch up .Thanks.
 one year ago
See more questions >>>
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.