Mechanical Engineering - Machine Design - Discussion
Discussion Forum : Machine Design - Section 10 (Q.No. 31)
31.
A spring of stiffness 1000 N/m is streteched initially by 100 mm from the undeformed position. The work required to stretch it by another 100 mm is
Discussion:
6 comments Page 1 of 1.
Baskar said:
1 year ago
However, if the question had asked for the work required to stretch the spring by a total of 20 cm (initial stretch + additional stretch), then the calculation would be:
W = (1/2) k (x2^2 - x1^2).
W = (1/2) x 1000 x (0.2^2 - 0.1^2).
W = (1/2) x 1000 x (0.04 - 0.01).
W = (1/2) x 1000 x 0.03.
W = 15 Nm.
W = (1/2) k (x2^2 - x1^2).
W = (1/2) x 1000 x (0.2^2 - 0.1^2).
W = (1/2) x 1000 x (0.04 - 0.01).
W = (1/2) x 1000 x 0.03.
W = 15 Nm.
(1)
Jitu said:
1 decade ago
F = k.x using this we get F = 100N force is required to deform it by 100mm.
Now the work done = F.s=100*.1 = 10N-m.
So, the correct answer should be C.
Now the work done = F.s=100*.1 = 10N-m.
So, the correct answer should be C.
Mithin said:
4 years ago
@Manoj Kumar Yadav.
How do we get the value of "d" in the equation "work done =Favg*d"?
Please explain about it.
How do we get the value of "d" in the equation "work done =Favg*d"?
Please explain about it.
Gaurav said:
1 decade ago
Potential energy stored is variable here so integrate (Kx.dx) from 0.1 to 0.2 m.
You will get 15 N-m answer.
You will get 15 N-m answer.
(1)
Manoj Kumar Yadav said:
5 years ago
F1=1000*.1=100N, F2=1000*.2=200N.
So, Favg=F1 +F2 /2 =100+200/2=150,
work done =Favg*d=150 * .1=15 ans.
So, Favg=F1 +F2 /2 =100+200/2=150,
work done =Favg*d=150 * .1=15 ans.
Ajaypal said:
9 years ago
Potential Energy = 1/2 * k[(0.2)^2 - (0.1)^2].
= 15 N-m.
= 15 N-m.
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