a force of 10 pounds is required to stretch a spring of 4 inches beyond it's natural length. assuming hooke's law applies, how much work is done in stretching the spring from its natural length to 6 inches beyond its natural length.
what i did so far:
F(x)=kx
10=k4
k=5/2
by hooke's law i need to find the integral of 5/2x right. but what are the bounds.
All you have to do is use the formula
Work = potential energy increase = (1/2) k x^2, where k is the spring constant that you computed (which is 2.5 lb/inch) and x = 6 inches.
The answer wil be in inch-lb. If you want it in foor-lb, you will have to divide by 12.
You could also do it by integration
Work - Integral of F dx = Integral of kx dx from x=0 to x=6. The answer is the same.
15/4...... (1/2)(2.5)(6^2)=45 in/lb converto to feet divide by 12... 15/4
To find the work done in stretching the spring from its natural length to 6 inches beyond its natural length, you can use the formula for potential energy increase in a spring, which is equal to the work done.
The formula is: Work = (1/2) k x^2, where k is the spring constant and x is the displacement from the natural length.
In this case, you have already calculated the spring constant k to be 5/2. So the formula becomes: Work = (1/2) (5/2) (x^2).
To find the work done in stretching the spring from its natural length (0 inches) to 6 inches beyond its natural length, you substitute x = 6 into the formula:
Work = (1/2) (5/2) (6^2)
= (1/2) (5/2) (36)
= (5/4) (36)
= 45 inch-lbs
So the work done in stretching the spring from its natural length to 6 inches beyond its natural length is 45 inch-lbs. If you want it in foot-lbs, you can divide by 12:
Work = 45 inch-lbs / 12
= 3.75 foot-lbs.
To compute the work done in stretching the spring from its natural length to 6 inches beyond its natural length, you can use the formula for potential energy increase, which is equal to the work done.
The formula is:
Work = potential energy increase = (1/2) k x^2
where k is the spring constant (which you calculated as 2.5 lb/inch) and x is the displacement from the natural length (which is 6 inches in this case).
Plugging in the values, we get:
Work = (1/2) * 2.5 * (6^2) = (1/2) * 2.5 * 36 = 18.75 inch-lb
If you want the answer in foot-lb, you can divide it by 12:
Work = 18.75 inch-lb / 12 = 1.5625 foot-lb
Therefore, the work done in stretching the spring from its natural length to 6 inches beyond its natural length is 18.75 inch-lb or approximately 1.5625 foot-lb.