A force of 10 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work is done in stretching it
from its natural length to 6 in. beyond its natural length?
1 answer:
Answer:
W = 15/4 ft . ib
Step-by-step explanation:
Force = 10ib
According to hooked law, f(x) = kx
x = 4inches = 4/12 ft
x= 1/3ft
f(x) = 1/3k
10 = 1/3k
k = 30 ib/ft
f(x) = 30x
Workdone = integral of f(x) with its limit
6 inches = 6/12 ft
= 1/2ft
W = integral(1/2 to 0) of 30x
W = 15x^2(1/2 to 0)
W = 15(1/2)^2 - 15(0)^2
W = 15(1/4) - 0
W = 15/4 ft. Ib
You might be interested in
Answer:
Answer: 15
Step-by-step explanation:
57.10
70.9
these are only the answers for the first two problems
Answer:
Frieda's work is correct.
Step-by-step explanation:
z+22= 64 can also be z= 64-22
64-22= 42
Meaning, z= 42
42+22-22= 64-22
42= 42
Answer:
$15 is the answers
Step-by-step explanation:
please mark me as brainlest