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: very too bad for you buddy study nexts time
Step-by-step explanation:
The first one is 810
The second is -120
Answer:
$1.65
Step-by-step explanation:
The answer is 164 I believe
This is easy to do because you can factor
example
6/2=3 because 6=2*3 so 6/2=3/1 times 2/2 or 3
2x^3+17x^2+23x-42 can be factored out to equal
(x-1)(x+6)(2x+7)
so [(x-1)(x+6)(2x+7)]/(2x+7)=[(x-1)(x+6)] times (2x+7)/(2x+7)=(x-1)(x+6)=x^2+5x-6
the answer is (x-1)(x+6) or x^2+5x-6
