It takes 5.28 J of work to stretch a Hookes Laq spring 9.82 cm from its unstressed length. How much extra work is required to stretch an additional 11.9 cm? anwser in J
Figure the Potential energy at 11.9, and at 9.82cm. The difference is the energy that stretched the spring.
Or, the calculus way..
Work= INT kxdx from .0982 to .119 m
i have nooooooooo clue sowwie
No problem, I can help you with that!
To solve this problem, we need to calculate the work done in stretching the spring an additional 11.9 cm. We can do this by finding the difference in potential energy of the spring at this extension compared to its extension at 9.82 cm.
First, let's find the potential energy at 11.9 cm:
Potential energy (PE) = (1/2)kx²
Where k is the spring constant and x is the displacement.
Next, we need to find the potential energy at 9.82 cm:
PE = (1/2)kx²
Now, subtract the potential energy at 9.82 cm from the potential energy at 11.9 cm to find the extra work required:
Extra work = PE at 11.9 cm - PE at 9.82 cm
Alternatively, you can also solve this using calculus:
We can express the work done in stretching the spring as the integral of the force over displacement. The force exerted by the spring is given by Hooke's Law: F = -kx, where k is the spring constant and x is the displacement.
The work done is then given by the integral:
Work = ∫ Fdx
To find the limits of integration, we need to calculate the displacement difference between 9.82 cm and 11.9 cm:
Δx = 11.9 cm - 9.82 cm
Now we can calculate the extra work required:
Work = ∫ Fdx from 9.82 cm to 11.9 cm
Finally, to get the answer in joules (J), you need to know the value of the spring constant (k). If the spring constant is given in the problem or known, you can substitute its value into the equations.
I hope this explanation helps!