Question

A force of 10 lb is required to hold a spring stretched 2 ft. beyond its natural length. How much work is done in stretching the spring from 2 ft. beyond its natural length to 5 ft. beyond its natural length

Answer 1

workdone =30.5J or 22.5 Ib.ft

Explanation:

From Hookes law , F=ke

k=F/e

F=10 Ib , e= 2f

K=10/2ft =5 Ib/ft

when e = 5ft

F=ke

F= 5 Ib/ft *5 ft

F=25 Ib

force required for stretching from 2ft to ft beyond its natural length equals =25-10 =15 Ib

Workdone in stretching a spring is = 1/2Fe

= 1/2* 15 Ib * 3ft

=22.5 Ibft

1 Ibft is equal to 1.35582joules.

22.5 Ibft= 22.5*1.35582 = 30.5059J

workdone =30.5J