Friday

February 27, 2015

February 27, 2015

Posted by **COFFEE** on Friday, June 29, 2007 at 11:00pm.

and this is what i did.. please check to see if i did it correctly.. thanks :)

f=27N

12cm to 15cm

f(x) = kx, k=spring constant

15-12cm = 3cm = 0.03cm

f(0.03) = 27N

0.03k = 27

k = 27/0.03 = 900

f(x) = 900x

w = integral of 0.03 to 0.13 for: 900x dx

w = 900*((x^2)/2)] from 0.03 to 0.13

w = 450[((0.13)^2)-((0.03)^2)]

w = 7.2J

Ok on k, 900 N/m

Ok on the next.

thanks :)

**Answer this Question**

**Related Questions**

work force - a force of 10 pounds is required to stretch a spring of 4 inches ...

Sorry, another math problem please - Hooke's Law asserts that the force required...

Physics - Hooke's law states that it takes a force equal to kΔx is required...

calculus - a spring exerts a force of 150 N when it is stretched 0.4 m beyond ...

Calculus - A spring has a natural length of 24 cm. If a 27-N force is required ...

Integral Calculus - A force of 4 pounds is required to hold a spring stretched 0...

phy - A string of natural length L extends to a new length L' under tensile ...

Math-Hooke's law - Using Hooke's Law (F=kx) determine: A) A weight of 5 pounds ...

Calculus - Work of 1 joules is done in stretching a spring from its natural ...

math - A spring with a mass of 4 kg has damping constant 28, and a force of 12 N...