dE= kdx
6J= k(.02m)
10J= k(.04m)
k= (6J)/(.02m)
k= (10J)/(.04m)
k= 300N/m
Now use the equation
E= 1/2 kx^2
E= 1/2 (300N/m) (xo)^2
xo= sqrt(12J/300N/m)
xo= 0.05m
The natural length of the spring is 0.05m.
If 6 J of work is needed to stretch a spring from 10cm to 12 cm and another 10 J is needed to stretch it from 12 cm to 14cm, what is the natural length of the spring?
This is a physics problem as much as a calculus one. Solve for the spring constant k first, then determine where the spring started from.
would k be 4?...then what next?
I think k is -5, but I'm not sure of the units. 10 J is needed to stretch it 2cm, so -5 something is needed per cm. The spring constant is usually expressed as a negative number when work is needed to stretch the spring.
It appears the spring actually did some work before work was performed on it to stretch to 12cm. Use 10 as the starting point. Solve
(2-x)*5 - 5*x = 6
and add x to 10
I think the answer is between 10 and 10.5
Work is related to the square of distance stretched
Force= kx
Energy= 1/2 kx^2 where x is the stretched distance. This is a second degree equation, not linear.
Let xo be the unstretche position.
Then the first set of data..
6 J= 1/2 k( .12-xo)^2 - 1/2 k (10-xo)^2
second set of data..
16J= 1/2 k (.14-xo)^2 -1/2 k (.10-xo)^2
You have two equations, two unknowns. In this problem, you are looking for xo.
you can also use differentials to work this.
Remember to change cm to m, to make it consistent with the joules energy (joule= newton*meter). Then k will be in newton/meter or joule/m^2)
1 answer
1 answer